Problem 3. (5 points) Initially, there is a cannon on every square of a chessboard. On each move, a cannon that threatens an odd number of other cannons can be removed from the board. What is the maximum number of cannons that can be removed from the board? (Two cannons threaten each other if they are in the same row or column of the board and there are no other cannons between them.)
Solution: We will prove that at most 59 cannons can be removed.

First, note that none of the cannons in the corner fields can be removed. Let's assume that the "corner" cannon from $a1$ is removed first. At the moment before its removal, it was threatening two cannons: one of those on line $a$ (there is at least $a8$) and one of those on line 1 (there is at least $h1$). Contradiction.

If only the corner cannons are left on the board, there is no field that is threatened by an odd number of cannons. Therefore, we cannot leave only the four corner cannons (otherwise, the last removed cannon would have been threatening 0 or 2 cannons). Therefore, at least 5 out of the 64 cannons will remain on the board. Here is how 59 cannons can be sequentially removed: $a7, a6, b6, a5, b5, c5, a4, b4, c4, d4, a3, b3$, $c3, d3, e3, a2, b2, c2, d2, e2, f2, b1, c1, d1, e1, f1, g1, c8, d8, e8, f8, g8, d7, e7, f7, g7$, $h7, e6, f6, g6, h6, f5, g5, h5, g4, h4, h3, h2, g2, g3, f3, f4, e4, e5, d5, d6, c6, c7, b7$.