Let $n$ and $k$ be positive integers such that $\frac{n}{2}<k \leq \frac{2 n}{3}$. We place $m$ pawns on an $n \times n$ chessboard, in such a way that no row or column contains a block of $k$ consecutive empty squares. What is the smallest integer $m$ for which this is possible?
Let's number the cells by pairs $(i, j)$, where $1 \leq i \leq n$ and $1 \leq j \leq n$. There is a natural configuration that meets the conditions of the problem and contains few pawns: place a pawn on the cell $(i, j)$ if and only if $i+j-1$ is divisible by $k$. For any pair $(i, j)$, $i+j-1$ is strictly less than $4k$, so the cells with pawns correspond to the values $k, 2k$, and $3k$ of $i+j-1$, and represent diagonal lines on the chessboard. Since $3k \leq 2n$, it is easy to count the number of pawns: it is $k + (2n - 2k) + (2n - 3k) = 4(n - k)$.

To prove that this construction is indeed optimal, let's make the following partition: we partition the chessboard into 9 rectangular regions $A, B, C, D, E, F, G, H, I$, such that the four corners $A, C, G$, and $I$ are squares with side length $n-k$. (Thus, $B, D, F$, and $H$ are rectangles with sides $n-k$ and $2k-n$.)

$$
\left(\begin{array}{lll}
A & B & C \\
D & E & F \\
G & H & I
\end{array}\right)
$$

The region $A \cup B$ consists of $n-k$ rows of $k$ cells. Each of these rows must contain a pawn, and therefore $A \cup B$ contains at least $n-k$ pawns. We reason similarly with the regions $C \cup F, I \cup H$, and $G \cup D$, which also contain at least $n-k$ pawns. This shows that the chessboard contains at least $4(n-k)$ pawns.