Example 8 Let $n$ be a fixed positive integer, consider an $n \times n$ square board, which is divided into $n^{2}$ unit squares, two squares on the board are called adjacent if they share a common edge.

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


Solve: For an $n \times n$ ($n \geqslant 2$ and $n$ is even) grid, color the cells in a "square" pattern with black and white, such that the outermost "square" contains $4n-4$ black cells, the second "square" contains $4n-12$ white cells, and so on, alternating black and white. It is easy to verify that there are $\frac{1}{2} n(n+2)$ black cells, and each marked cell is adjacent to exactly two black cells (regardless of whether the marked cell is black or white). For any $N$ that meets the problem's requirements, in the black cells adjacent to each marked cell, each black cell must appear at least once; otherwise, if there exists a black cell not adjacent to any marked cell, this would contradict the problem's requirement that each cell (especially black cells) must be adjacent to at least one marked cell. Therefore, 
$2N \geqslant \frac{1}{2} n(n+2)$, which means $N \geqslant \frac{1}{4} n(n+2)$ ($n \geqslant 2$ and $n$ is even).

On the other hand, label the cells of the outermost "square" of the $n \times n$ grid, starting from the first cell of the first row, with the numbers $1,1,2,2, \cdots, 2n-2,2n-2$; then remove the outer two layers of "square" cells and re-label the cells of the new outermost "square" starting from the first cell of the first row with the numbers $1,1,2,2, \cdots, 2n-10,2n-10$; continue this process until all cells are labeled. Mark the cells labeled with odd numbers, and there will be $\frac{1}{4} n(n+2)$ marked cells, which can be easily verified to meet the requirements. Therefore, the minimum value of $N$ is $\frac{1}{4} n(n+2)$.