How is this process not quasi-static yet reversible?

In the context of equilibrium thermodynamics, when we say that a process is reversible, we usually mean that the process is well-described by a continuous path in thermodynamic state space (namely it is quasi-static) and that the system can be taken along the same path in thermodynamic state space, but in the reverse order. Since in the example you gave, the system is only in equilibrium in the initial and final states, it doesn't make sense to talk about reversibility.

It's true that the system is going from one equilibrium point (point in thermodynamic state space) to another and then back again, but one cannot speak of the system "retracing" its thermodynamic steps so-to-speak because from the perspective of equilibrium thermodynamics, these intermediate steps are not well-defined macroscopic states.

In the process you describe the system won't necessarily return to its original state.

Suppose you instantly remove the weight and lift up the piston so the gas expands irreversibly to it's new equilibrium volume. The gas does no work in expanding so its temperature doesn't change - all that happens is that the pressure falls.

Now compress the gas back to it's original volume reversibly by lowering the weight back on very gradually. Now you are doing work on the gas so its temperature increases. When you eventually release the weight you'll find the gas temperature and volume are different to the initial state.

If you do the cycle reversibly you'll find that the work done by the gas as it expands is the same as the work done on the gas as it's compressed, and it will return to its initial state. However this isn't the case if the expansion/compression is not revesible. This is because work is not a function of state.

If I put the weight back on the piston, the system will again achieve its initial state.

No, it won't. In the end, the pressure will be the same, but the temperature and therefore the volume will be higher.

Firstly, in a real system there will be friction due to gas viscosity and piston/cylinder interaction.

But even in an ideal system, after the first weight is removed, the gas expands against only the pressure of the remaining weight(rather than the gas's own higher pressure) and therefore does less work than it otherwise could. $Delta U_1 = -W_1$. When the weight is replaced, the compression is against the gas's own pressure, and more work would be required to return the piston to its orginial position than the gas orginially did. $Delta U_2$ would need to be of greater magnitude than $Delta U_1$ to get back to the starting pistion position.

see section 3.3.2 and 3.3.3 of the following reference:

http://sites.tufts.edu/andrewrosen/files/2012/11/Thermo-Review.pdf

The process is obviously quasi static if dynamics are ignored and as long as no friction is assumed also reversible. Perhaps the key is to consider what "removing the weight" actually means. If the weight is removed by a vertical action from $h_0$ to $h_1$ (heights, where the indices mean equilibrium positions) and you do it really slowly (dynamic effect by moving masses neglected), the process is quasi static.

If however the weight is removed quickly or even instantanously (eg. by removing the weight by horizontal action at $h_0$) the system will oscillate stationarily around $h_1$. The energy stored in the oscillation would be the difference between the potential energy $mg(h_1-h_0)$ and isentropic volume change work $int_0^1 pdV$, where $m$ is the removed mass, $p=p_0(V_0/V)^kappa$ and $kappa=c_p/c_v$. While obviously in this situation still everything is reversable, it's also still quasi static with respect to internal of the closed system where no non-equilibrium processes take place, besides some local effects which average out. In other words for any position $h$ of the piston the internal state is given independently of circumstances outside the system.

The other answers showed that the energy of the final system is greater than the energy of the initial system, because you are heating the gas. Here's another way to see this:

In the first step, you remove the weight at height $h_1$. In the second step you replace it at a higher height, $h_2$. In between these steps you need to bring the weight up from $h_1$ to $h_2$, which takes work. The environment will lose energy in performing this work, so the system must gain energy to compensate. One might object that you could just even things out in the end by transferring energy back from the system to the environment. However, the environment lost energy in the form of work, and the system gained energy in the form of heat, so the net energy transfer is irreversible.

It is just not a reversible process.

Calculations for the irreversible sequence

Say the piston has an area $A$, the two masses are $m$ and $m'$, and the initial state of the gas is characterized by:

$$V = V_0,$$ $$P = P_0 = P'_e = (m + m') g / A,$$ $$T = T_0.$$

with $P'_e$ the external pressure applied to the piston when both masses $m$ and $m'$ are present.

When the mass $m'$ is removed, the gas, at initial pressure $P_0$ expands against the new external pressure $P_e = m g / A < P_0$, doing a work $W = -P_e Delta V$ until it reaches the new state: $$V = V_f,$$ $$P = P_f = P_e,$$ $$T = T_f.$$

When the mass $m'$ is repositioned, the gas is compressed by the new external pressure $P_e'$, doing a work $W' = - P_e' Delta V'$.

At the end of the process, the internal energy of the gas has changed by the amount $$Delta U = W + W' = - P_e Delta V - P'_e Delta V'.$$ So even if $Delta V'$ were equal to $Delta V$ the system is not brought back to the same state.

The crucial point here is that during the expansion and compression phases, there is a discontinuity between the internal and external pressure, and therefore an asymmetry between the expansion and the compression. This discontinuity is the reason why it cannot be considered as a quasi-static process. In fact, the term "quasi-static process" is a misnomer because what it designates it is not a process at all but a range of constrained equilibrium states. Giving this range of constrained equilibrium states determines nothing about the various actual possible processes that could navigate the system from one equilibrium state to the next.

Scenario of a reversible sequence (reversible cycle)

For the process to be reversible, the external pressure applied to the piston should always equal the pressure of the gas $P(t) = P_e(t)$ all along the transformation, with for instance $P(t)= ntext{R}T(t)/V(t)$ if it were a perfect gas. Such a process would require to "remove" the mass $m'$ very progressively.

In such a process the work done by the gas would be exactly the opposite during the compression as it was during the expansion, and the gas would then be brought back to its initial state.

Entropy increase

One might then wonder how the entropy has increased during the first irreversible sequence, since $Q = 0$ all along the process which is assumed to be adiabatic... There are two ways to analyze the situation:

1. The first way is to recognize, as explained in the answer by Nathaniel, that for the system to reach a static equilibrium, there must be friction happening which stabilizes the piston at each new position. If there were no such friction, the piston would keep oscillating after the mass $m'$ had been removed or after it had been repositioned.

2. The second way is to be aware that the relation $text{d}S = delta Q/T$ only applies to reversible processes. In the irreversible process, the entropy increases due to the pressure discontinuity in the composite system ($P_0 neq P_e$). To compute the resulting increase in entropy, one would need to find a reversible process that bring the system (and its surrounding) to exactly the same final state, and evaluate $text{d}S = delta Q/T$ for the gas along this reversible process. Obviously, such a reversible process would not be adiabatic, but it would be computable.

Are you sure the gas is really going out of equilibrium?

Quote from General Physics, by L. D. Landau and others (the link requires login, to "borrow" the book):

The context is the expansion of a gas in a thermally isolated cylindrical vessel with a piston.

Here, "sufficiently slowly" means, therefore, so slowly that the gas is able to establish thermal equilibrium corresponding to every instantaneous position of the piston...Analysis shows that the condition would not be fulfilled only if the rate of movement of the piston were comparable with the velocity of sound in the gas."

Here's a similar quote from Pippard, where by fluid he means a gas or liquid:

Usually in practice, however, a fluid requires very little time to equalize its pressure, and the change can be carried out quite quickly without provoking a significant departure from equilibrium.