Example 6 Let $n$ be a fixed positive even number. Consider an $n \times n$ square board, which is divided into $n^{2}$ unit square cells. Two different cells on the board are called adjacent if they share a common edge. Mark $N$ unit square cells on the board so that every cell on the board (marked or unmarked) is adjacent to at least one marked cell. Determine the minimum value of $N$.
(40th IMO)
Solution: Let $n=2k$.
First, color the square board in a checkerboard pattern like a chessboard. Let $f(n)$ be the minimum value of $N$ sought, $f_{a}(n)$ 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 symmetrical, hence
$$
f_{a}(n)=f_{b}(n), f(n)=f_{a}(n)+f_{b}(n) \text {. }
$$

For convenience, place the "board" horizontally along the longest diagonal of black squares, then the number of black squares in each row is $2,4, \cdots, 2 k, \cdots, 4,2$.

Below the row containing $4 i-2$ black squares, mark the white squares in odd positions.

When the row is above the diagonal, a total of $2 i$ white squares are marked, as shown in Figure 3; when the row is below the diagonal, a total of $2 i-1$ white squares are marked, as shown in Figure 4.
Thus, the number of white squares marked is
$2+4+\cdots+k+\cdots+3+1=\frac{k(k+1)}{2}$ (squares).
It is easy to see that at this point each black square is adjacent to 1 marked white square, so we have $f_{a}(n) \leqslant \frac{k(k+1)}{2}$.

Consider these $\frac{k(k+1)}{2}$ marked white squares. None of them have adjacent common black squares, so at least $\frac{k(k+1)}{2}$ black squares need to be marked to ensure that each of these white squares has a marked black square adjacent to it.
Thus, $f_{b}(n) \geqslant \frac{k(k+1)}{2}$.
Therefore, $f_{a}(n)=f_{b}(n)=\frac{k(k+1)}{2}$.
Hence, $f(n)=k(k+1)$.