4. Let $n$ and $k$ be positive integers, satisfying $\frac{1}{2} n < k \leqslant \frac{2}{3} n$. Find the smallest number $m$, such that placing each of $m$ "pawns" in one square of an $n \times n$ chessboard ensures that no row or column contains $k$ consecutive squares without a "pawn".
Solution: If there does not exist a $k \times 1$ or $1 \times k$ block of squares without a "pawn" on the chessboard, then the placement of "pawns" on the chessboard is called "good". Let the rows and columns be from the 0th row to the $(n-1)$th row and from the 0th column to the $(n-1)$th column, respectively.

An ideal "good" placement is as follows: Place a "pawn" at $(i, j)$, where $i, j$ are the $i$th row and $j$th column, respectively, such that $i+j+1$ is divisible by $k$. Since $n > 0$, this division is reasonable.

Assume that in the rectangular region $B$, there are $b$ rows without "pawns", $H$ has $h$ rows without "pawns", $D$ has $d$ columns without "pawns", and $F$ has $f$ columns without "pawns". Take any one of the $b$ rows without pawns in $B$ and extend it to the left and right across the entire chessboard. Then, the part of this row extended into $A$ must contain at least 1 "pawn", otherwise the placement is not "good". Similarly, the part of this row in $C$ must also contain at least 1 "pawn". Mark the squares in $A$ and $C$ where these two "pawns" are located.

Similarly, the same conclusion applies to the $h$ rows without "pawns" in $H$. The same conclusion also applies to the $d$ columns without "pawns" in $D$ and the $f$ columns without "pawns" in $F$. Therefore, in $A \cup C \cup G \cup I$, a total of $2(b+h+d+f)$ marks are made on the squares with "pawns", and each square is marked at most twice. Thus, the number of squares with marks is at least $b+h+d+f$, and each of these squares contains 1 "pawn". Therefore, there are at least $b+h+d+f$ "pawns" in $A \cup C \cup G \cup I$.

Since $B$ has at least $n-k-b$ "pawns", $H$ has at least $n-k-h$ "pawns", $D$ has at least $n-k-d$ "pawns", and $F$ has at least $n-k-f$ "pawns", it follows that $m \geqslant 4(n-k)$. That is, $4(n-k)$ is the minimum value.