jmc

The vertices of the triangles are limited to a $4\times 4$ grid, with $16$ points total. Every triangle is determined by $3$ points chosen from these $16$ for a total of $\left(\genfrac{}{}{0ex}{}{16}{3}\right)=560$. However, triangles formed by collinear points do not have positive area. For each column or row, there are $\left(\genfrac{}{}{0ex}{}{4}{3}\right)=4$ such degenerate triangles. There are $8$ total columns and rows, contributing $32$ invalid triangles. There are also $4$ for both of the diagonals and $1$ for each of the $4$ shorter diagonals. There are a total of $32+8+4=44$ invalid triangles counted in the $560$, so the answer is $560-44=\boxed{516}$.