Why Hamiltonian of gravity is zero?

I won't give a full derivation -- for that I will point to the literature (or really I'll point to my comment where I gave a ref to the literature), where this calculation has been done many times.

Instead I'll give an outline.

We start with the Einstein-Hilbert action$^star$

begin{equation} S = frac{1}{16pi G} int {rm d}^4 x sqrt{-g} R end{equation}

We then define the lapse function $N=(-g^{00})^{-1/2}$, shift vector, $N_i = g_{i0}$, and spatial metric $h_{ij} = g_{ij}$. (The index heights are crucially important here, and please note I am choosing the fastest way through the algebra intentionally so that I do not take it on myself to explain every step).

We also can derive a canonical momentum $pi^{ij}$, which comes from differentiating the action wrt time. (If you work through the Dirac procedure, you will see the lapse and shift do not have an associated momentum, and instead are Lagrange multipliers for first class constraints).

Then the Lagranian can be written in the canonical form

begin{equation} mathcal{L} = - g_{ij} dot{pi}^{ij} - N H - N_i P^i + ({rm total derivative}) end{equation}

where the Hamiltonian constraint $H$ associated with the lapse $N$ is begin{equation} H = sqrt{h} left(^{(3)}R+h^{-1} (frac{1}{2}pi^2 - pi^{ij} pi_{ij})right) end{equation} and the momentum constraints $P^i$ associated with the shifts $N_i$ are begin{equation} P^i = -2 nabla_j pi^{ij} end{equation}

where $^{(3)}R$ and $nabla_j$ are the Ricci scalar and covariant derivative associated with the spatial metric $h_{ij}$, respectively.

As I said in the comments, the thing you specifically asked about, namely the fact that the Hamiltonian is zero on the constraint surface, is not special to gravity and instead is related to diffeomorphism invariance, which is also exhibited by the free relativistic particle (but easier to understand in that context). However, this is the canonical formulation of gravity.

$^star$ strictly speaking one needs to supplement the action with the Gibbons-Hawking-York boundary terms.