From rafayaashary1 rafayaashary1 wrote: First note that each odd-length diagonal must have at least have one "hole" (absence of stone). Furthermore, we can easily check that each cell is in at most one odd-length diagonal, and that there are $24$ total odd-length diagonals. Thus we have at most $120$ stones. I claim that this bound is tight. Indeed, consider the grid with every cell not on a "main diagonal" filled. It is easily checked that said configuration satisfies the conditions of the problem, and we conclude that the answer is $\boxed{120}\ \blacksquare$

We notice that in our board,two odd length diagonals dont intersect. Each such diagonal contains an unfilled cell.Therefore there are at least 24 unfilled and atmost 120 stones.