smc

Let the side length of the square be $s$. If $s$ is even, there are $s$ black tiles on each of the two diagonals, so the total number is $2s$. If $s$ is odd, we double-count the middle square which is on both diagonals, so the number is instead $2s-1$. In this case, since $101$ is odd, we clearly must have the second case, so $2s-1=101⟹2s=102⟹s=51$, and indeed $s$ is odd as we expected. Thus the overall number of tiles is $51^2=2601$. $\boxed{2601}$