Solution:

We use Pick's Theorem, which states that in a lattice polygon with $I$ lattice points in its interior and $B$ lattice points on its boundary, the area is $I + B/2 - 1$. Also, call a point center if it is of the form $\left(x+\frac{1}{2}, y+\frac{1}{2}\right)$ for integers $x$ and $y$.

The key observation is the following: suppose we draw in the center points, rotate $45^{\circ}$ degrees about the origin and scale up by $\sqrt{2}$. Then, the area of the triangle goes to $2K$, and the set of old lattice points and center points becomes a lattice. Hence, we can also apply Pick's theorem to this new lattice.

Let the area of the original triangle be $K$, let $I_1$ and $B_1$ be the number of interior lattice points and boundary lattice points, respectively. Let $I_c$ and $B_c$ be the number of interior and boundary points that are center points in the original triangle. Finally, let $I_2$ and $B_2$ be the number of interior and boundary points that are either lattice points or center points in the new triangle. By Pick's Theorem on both lattices,
$$\begin{aligned} K &= I_1 + B_1/2 - 1 \\ 2K &= I_2 + B_2/2 - 1 \\ \Longrightarrow (I_2 - I_1) &= K - \frac{B_1 - B_2}{2} \\ \Longrightarrow I_c &= K - \frac{B_c}{2}. \end{aligned}$$
One can compute that the area is $31500$. The number of center points that lie on $AB$, $BC$, and $CA$ are $0$, $10$, and $30$, respectively. Thus, the final answer is $31500 - \frac{0 + 10 + 30}{2} = 31480$.