The Number of distinctly oriented square types with a $k\times k$ envelope in a grid is $k$ and the number of such envelopes in the $n\times n$ array of points is $(n-k)^2$. So for an $n\times n$ array of points, Gastón can choose the following number of squares: $$P=\sum_{k=1}^{n-1}k(n-k)^2=\sum_{k=1}^{n-1}(k^3-2nk^2+n^2k)=\frac{n^2(n-1)^2}{4}-\frac{n^2(n-1)(2n-1)}{3}+\frac{n^3(n-1)}{2}=\frac{n^2(n+1)(n-1)}{12}$$ In this case, we have $n=5$, so $P=\frac{5^2\cdot 6\cdot 4}{12}=50$ choices for Gastón.

I don't think this is correct. It's possible to have squares that are "tilted", basically the ones whose sides are not aligned with the axes.

Do you have the official PDF for this olympiad?