None of the rooks standing in the corner cells can be taken. Indeed, at the moment when the first of them is taken, it is attacked by exactly two rooks. Suppose it was possible to leave only the four corner rooks. Then the last rook taken was attacked by either two rooks (if it stood on the edge of the board) or none. Contradiction. Therefore, at least five rooks will remain.

We will show how to take 59 rooks. First, we will take all the rooks from the 8th rank, except for three - a8, b8, and h8, then repeat the same with the 7th, 6th, ..., 3rd, and 1st ranks (skipping the 2nd). Now we will take the rooks from a3 to a7 on the 1st file, and from h3 to h7 on the last file. Next, we will take the rooks b1 and a2. Finally, we will sequentially take the rooks from b8 to b3 and from g2 to c2.

## Answer

59 rooks.

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Shapovalov $A$.

On each square of the chessboard, there is a rook at the beginning. In each move, you can remove a rook that attacks an odd number of rooks. What is the maximum number of rooks that can be removed? (Rooks attack each other if they stand on the same vertical or horizontal line and there are no other rooks between them.)