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

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


Solution: Let $n=2k$. 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, $f_{w}(n)$ 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 "chessboard" is symmetrical. Therefore,
$$
\begin{array}{l}
f_{w}(n)=f_{b}(n), \\
f(n)=f_{w}(n)+f_{b}(n) .
\end{array}
$$

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

Below the row with $4i-2$ black squares, mark the white squares in the odd positions. When the row is above the diagonal, there are $2i$ marked white squares (see Figure 1); when the row is below the diagonal, there are $2i-1$ marked white squares (see Figure 2). 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 at this point, each black square is adjacent to a marked white square, so we have
$$
f_{w}(n) \leqslant \frac{k(k+1)}{2} .
$$

Consider these $\frac{k(k+1)}{2}$ marked white squares. No two of them have a common 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,
$$
f_{b}(n) \geqslant \frac{k(k+1)}{2} \text{. }
$$

Hence, $f_{w}(n)=f_{b}(n)=\frac{k(k+1)}{2}$.
Therefore, $f(n)=k(k+1)$.