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 $2 K$, 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, $$\begin{aligned} K & =I_{1}+B_{1} / 2-1 \\ 2 K & =I_{2}+B_{2} / 2-1 \\ \Longrightarrow\left(I_{2}-I_{1}\right) & =K-\frac{B_{1}-B_{2}}{2} \\ \Longrightarrow I_{c} & =K-\frac{B_{c}}{2} \end{aligned}$$ One can compute that the area is 31500. The number of center points that lie on on $A B, B C$, and $C A$ are 0,10, and 30, respectively. Thus, the final answer is $31500-\frac{0+10+30}{2}=31480$.