24*. On an $8 \times 8$ chessboard, the squares are colored in two colors in a checkerboard pattern. A token moves on the black squares, and in one move, it can transition to any diagonally adjacent square. What is the minimum number of moves required to visit all the black squares?

## 9th grade
81.24. We will paint all black cells below the diagonal connecting the top left and bottom right corners of the board in red and blue as follows: if the sum of the row number and column number (numbers are counted from the bottom and left) is divisible by four, we paint the cell red; if it is not divisible by four, we paint it blue. Now we will reflect this coloring symmetrically relative to the aforementioned diagonal and paint all cells above it. Each move, except for moves crossing the diagonal, the piece moves from a blue cell to a red one and vice versa. Since there are eight more blue cells than red ones, and there are only four moves from a blue cell to a blue cell, the piece must make at least three moves to cells it has already visited. Therefore, the answer: the minimum number of moves is \(31+3=34\).