C50 (40-3, Belarus) Let $n$ be a fixed positive even number. Consider an $n \times n$ square board, which is divided into $n^{2}$ unit squares. Two different squares on the board are called adjacent if they share a common edge.

Mark $N$ unit squares on the board so that every square on the board (marked or unmarked) is adjacent to at least one marked square.
Determine the minimum value of $N$.


Let $n=2 k$. First, color the square board in a chessboard pattern with alternating black and white squares. Let $f(n)$ be the minimum value of $N$ sought, $f_{\omega}(n)$ be the minimum number of white squares that must be marked so that every black square has a marked white square adjacent to it. Similarly, define $f_{b}(n)$ as the minimum number of black squares that must be marked so that every white square has a marked black square adjacent to it. Since $n$ is even, the "board" is symmetric. Therefore,
$$
\begin{array}{c}
f_{\omega}(n)=f_{b}(n), \\
f(n)=f_{\omega}(n)+f_{b}(n) .
\end{array}
$$

For convenience, place the "board" with the longest diagonal of black squares horizontally, so the number of black squares in each row is $2,4, \cdots, 2 k, \cdots, 4,2$.
In the row containing $4 i-2$ black squares,

Mark the white squares in odd positions. When the row is above the diagonal, there are $2 i$ marked white squares (as shown in Figure 4-112); when the row is below the diagonal, there are $2 i-1$ marked white squares (as shown in Figure 4-113). Therefore, the total number of marked white squares is
$$
2+4+\cdots+k+\cdots+3+1=\frac{k(k+1)}{2} \text { (squares). }
$$

It is easy to see that each black square is adjacent to a marked white square, so we have
$$
f_{\omega}(n) \leqslant \frac{k(k+1)}{2} \text {. }
$$

Consider these $\frac{k(k+1)}{2}$ marked white squares; no two of them share an adjacent black square. Therefore, at least $\frac{k(k+1)}{2}$ black squares must be marked to ensure that each of these white squares has a marked black square adjacent to it, thus
$$
f_{b}(n) \geqslant \frac{k(k+1)}{2} .
$$

From (1), (3), and (4), we get
$$
f_{\omega}(n)=f_{b}(n)=\frac{k(k+1)}{2} .
$$

From (2) and (5), we get $f(n)=k(k+1)=\frac{n(n+2)}{4}$.