Problem 3. On each cell of a $10 \times 10$ board, a pawn is placed. We can choose a diagonal (a diagonal consists of all cells whose centers lie on a straight line that makes an angle of $45^{\circ}$ with the sides of the board), which contains an even number of pawns, and take one arbitrary pawn from it. How many pawns can be taken from the board in this way?
Solution. We will call a diagonal even (odd) if it contains an even (odd) number of pawns. Initially, the board has 20 odd diagonals. Let's consider how the number of these diagonals changes when a pawn is removed from the board (it lies on at least one even diagonal). If the pawn lies on one even and one odd diagonal, removing it does not change the number of odd diagonals. If the pawn lies on two even diagonals, removing it increases the number of odd diagonals by 2. Therefore, at any moment, there are at least 20 odd diagonals, i.e., at least 10 pawns will remain on the board.

Let's denote the cells $(i ; j)$, where $0 \leq i, j \leq 9$. We will show how to remove all pawns from the cells for which $i+j$ is odd, except for $(1 ; 0)$, $(3 ; 0)$, $(5 ; 0)$, $(7 ; 0)$, and $(9 ; 0)$. We sequentially remove the pawns from the cells:

| 0. | $(0 ; 1)$ | $(0 ; 3)$ | $(0 ; 5)$ | $(0 ; 7)$ | $(0 ; 9)$ | 1. | $(1 ; 2)$ | $(1 ; 4)$ | $(1 ; 6)$ | $(1 ; 8)$ |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| 2. | $(2 ; 1)$ | $(2 ; 3)$ | $(2 ; 5)$ | $(2 ; 7)$ | $(2 ; 9)$ | 3. | $(3 ; 2)$ | $(3 ; 4)$ | $(3 ; 6)$ | $(3 ; 8)$ |
| 4. | $(4 ; 1)$ | $(4 ; 3)$ | $(4 ; 5)$ | $(4 ; 7)$ | $(4 ; 9)$ | 5. | $(5 ; 2)$ | $(5 ; 4)$ | $(5 ; 6)$ | $(5 ; 8)$ |
| 6. | $(6 ; 1)$ | $(6 ; 3)$ | $(6 ; 5)$ | $(6 ; 7)$ | $(6 ; 9)$ | 7. | $(7 ; 2)$ | $(7 ; 4)$ | $(7 ; 6)$ | $(7 ; 8)$ |
| 8. | $(8 ; 1)$ | $(8 ; 3)$ | $(8 ; 5)$ | $(8 ; 7)$ | $(8 ; 9)$ | 9. | $(9 ; 2)$ | $(9 ; 4)$ | $(9 ; 6)$ | $(9 ; 8)$ |

Similarly, all pawns from the cells for which $i+j$ is even, except for $(0 ; 0)$, $(2 ; 0)$, $(4 ; 0)$, $(6 ; 0)$, and $(8 ; 0)$, are removed from the board. Thus, only the pawns from the first row remain on the board.