the maximum number is $21$. it's obvious that you cannot cut any of the border squares. in the $7\times 7$ square left at the middle, dissect it into four $3\times 4$, rectangles, from each small rectangle, you can cut at most $5$ squares + one square in the middle....

i find an arrangement with 32. this is an example with 64 cut. consider a square 8 by 8 in top- left of the grid and cut the table from the all diagonals which are parallel to main diagonal of the grid . so the answer is at least 32.

There's an old question which's term is that no two diagonals that have a common point. and the answer is 16 when n=5, 29 when n=7, 46 when n=9. (n means the $n\times n$ board)