. Pawns and rooks are placed on a $2019 \times 2019$ chessboard, with at most one piece on each of the $2019^{2}$ squares. A rook can see another rook if they are in the same row or column and all squares between them are empty. What is the maximal number $p$ for which $p$ pawns and $p+2019$ rooks can be placed on the chessboard in such a way that no two rooks can see each other?
Answer: the maximal $p$ equals $1009^{2}$.

Write $n=2019$ and $k=1009$; then $n=2 k+1$. We first show that we can place $k^{2}$ pawns and $n+k^{2}$ rooks. Each cell of the chess board has coordinates $(x, y)$ with $1 \leqslant x, y \leqslant n$. We colour each cell black or white depending on whether $x+y$ is even or odd.

Let $A$ be cell $(1, k+1), B$ be cell $(k+1,1), C$ be cell $(2 k+1, k+1)$ and $D$ be cell $(k+1,2 k+1)$, and consider the skew square $A B C D$. We place rooks on the cells of this square which have the same colour as $A$, and we place pawns on the other cells of this square. In this way, no rook can see another rook. Now we have placed $p=k^{2}$ pawns and $(k+1)^{2}=k^{2}+(2 k+1)=p+n$ rooks.

Now we show that we can not place more pawns. Observe that in every row the number of rooks exceeds the number of pawns by at most 1 , since there has to be a pawn between every two neighbouring rooks. So the total number of rooks exceeds the number of pawns by at most $n$. On the other hand we are to place $p$ pawns and $p+n$ rooks, so the number of rooks in every row exceeds the number of pawns by exactly 1 . This means that the rooks and pawns alternate, with rooks at the two ends. For the columns the same holds.

Consider the $\ell$-th row. Let $a$ be the number of pawns in this row and let $b$ be the number of pawns above the $\ell$-th row. For all these pawns, a rook must be somewhere above it. Counting the rooks directly above these $a+b$ pawns, we conclude that there must be at least $a+b$ rooks in the first $\ell-1$ rows. In every row the number of rooks exceeds the number of pawns by 1 , so in these first $\ell-1$ rows we have at least $a+b-(\ell-1)$ pawns. So $b \geqslant a+b-(\ell-1)$, yielding $a \leqslant \ell-1$. We conclude that the $\ell$-th row contains at most $\ell-1$ pawns. The same holds for the $\ell$-th row counted from below (row $(n+1)-\ell$ ): also in this row, there are at most $\ell-1$ pawns. As $n=2 k+1$, the maximal number $p$ is

$$
\sum_{\ell=1}^{k}(\ell-1)+\sum_{\ell=1}^{k+1}(\ell-1)=k+2 \cdot \frac{1}{2} k(k-1)=k+k(k-1)=k^{2}
$$