Is this a kinematics paradox?

This system seems counter-intuitive at first, but can be nicely explained once we analyze the torques acting on each of the gears. For convenience, from now on, I'll be referring to the blue gear on the left as $A$, blue gear on the right as $B$, orange gear on the left as $C$ and the orange gear on the right as $D$.

Now imagine rotating the orange shaft, and let's for a moment, forget the blue shaft. As you rotate the orange shaft, you'd see that a point on $C$'s boundary rotates at a speed $omega R$, and a point on $D$'s boundary rotates at a speed $2omega R$. The motion is similar for the blue shaft (in the absence of orange shaft) as well. Now when we put both of them together, the scenario unexpectedly changes.

Let's say we are applying a torque to rotate shaft 2 and the shaft 1 at that moment has no exterrnal torque acting on it. Once the shaft 1 starts moving, there are two possible scenarios (before I explain them, you should know that no matter how precisely you engineer these gears, there will always be some offset between both the pairs of touching gears ($AC$ and $BD$), thus one pair will come in contactbefore the other):

• $D$ touches $B$ before $C$ touches $A$ : In this case, since initially only $D$ and $B$ are touching, thus $B$'s surface acquires a velocity equal to $D$'s surface, and the initial angular velocity of shaft 1 becomes $2omega$ (here $D$ needs to push $B$ down to make it move). However, soon after the start, $C$ also starts touching $A$, but since $A$'s velocity before moving would have been $4omega R$, and $C$'s would have been $omega R$, thus they would not be able to move together after they touch each other (here $C$ stops $A$ from moving by pushing $A$ up). So finally, we would have $D$ applying a downward force on $B$, and $A$ applying a downward force on $C$. So the net torque on shaft 1 due to shaft 2 is in the north direction (see here for more on how to determine the direction of torque). This torque is balanced by the torque provided by the hinge to which the shaft 2 is attached.

• $C$ touches $A$ before $D$ touches $B$ : This scenario is extremely similar, so I leave it up to you to analyze it yourself. After analysis, we can see that the torque on shaft 1 due to shaft 2 is in the south direction, and again, this torque is balanced by the torque provided by the hinge to which the shaft 2 is attached.

What would happen if I apply a huge force to rotate shaft 2? Would the assembly still not move?

Well, if we are dealing with real objects, then at a certain strength of the torque/force you apply, the resulting torque on shaft 1 will exceed the maximum amount of torque that the hinge can provide, and eventually, the hinge would break apart and the shaft 1 would get displaced from its location. In ideal unbreakable situations, the system will remain static no matter what you do.

It's not a paradox at all: it just means that this setup of gears won't turn.

In fact, there's a related "paradox" that is extremely common in real life. Just replace your orange piece with the ground, and tilt the blue piece a bit, and you have a crude diagram of a car turning. The exact same reasoning you gave would apparently prove that cars can't turn without one of the wheels slipping on the ground.

This logic is perfectly correct if the car axles as rigid objects, which is why toy cars can't turn. Of course, real cars can turn, because the axle is broken with a differential, which allows the wheels to turn at different angular speeds.

By examining the gears on the left we can see that for every revolution of 1, 2 must undergo two revolutions. By examining the gears on the right we can see that for every revolution of 1, 2 must undergo half a revolution. This is not possible, so the gears will not turn.

It seems to me that what you see is what you get here - that arrangement of gears would lock up and the shafts would not turn.

In general, if a big gear drives a small gear then the smaller gear will rotate faster, and vice-versa. If you try to do both at once, then your effort will simply go into applying torsion to the secondary shaft, which will be trying to rotate at two different rates at the locations of the first and second gear.