7・119 Let $n$ be a positive even number. Consider an $n \times n$ grid. If two different cells share 1 edge, they are called adjacent. On this grid, mark $m$ cells so that each cell on the board is adjacent to at least 1 marked cell. Find the minimum value of the number of marked cells $m$.


[Solution] Color the grid board in a checkerboard pattern with black and white squares, and consider all the black squares. Clearly, the neighboring squares of black squares are all white squares. Therefore, we only need to consider the minimum number of white squares that need to be marked to ensure that each black square has at least one marked adjacent square.
Let $n=2 k, k \in \mathbb{N}$. Tilt the grid board and take the squares marked with circles in Figure (a) as the marked squares.
It is easy to see that each black square on the grid board is adjacent to exactly one marked square. When $k$ is even, the number of marked squares is
$$
\begin{array}{r}
2+4+\cdots+k+(k-1) \\
+\cdots+3+1=\frac{1}{2} k(k+1) ;
\end{array}
$$

When $k$ is odd, the number of marked squares is
$$
\begin{array}{r}
2+4+\cdots+(k-1)+k \\
+\cdots+3+1=\frac{1}{2} k(k+1) .
\end{array}
$$

This shows that taking $\frac{1}{2} k(k+1)$ marked white squares can ensure that each black square is adjacent to at least one marked square. Similarly, taking $\frac{1}{2} k(k+1)$ marked black squares can ensure that each white square is adjacent to at least one marked square, so the minimum number of marked squares $m$ is no more than $k(k+1)$.

On the other hand, consider the $\frac{1}{2} k(k+1)$ black squares marked with crosses in Figure (b). It is easy to see that they do not share any common adjacent squares. In other words, each white square can only be adjacent to one of these crossed black squares. Therefore, to meet the requirement, at least $\frac{1}{2} k(k+1)$ white squares need to be marked. Similarly, at least $\frac{1}{2} k(k+1)$ black squares need to be marked. Therefore, to meet the requirement, at least $k(k+1)$ squares need to be marked.

In summary, the minimum number of marked squares $m$ is $k(k+1)=\frac{1}{4} n(n+2)$.