【Question 24】
In the figure, color some cells so that any $3 \times 3$ square within the figure contains exactly 4 colored cells. How many cells need to be colored at a minimum?
【Analysis and Solution】
The problem of extremum.
First, we consider the 4 $3 \times 3$ squares in Figure 1;
To ensure that each of these 4 $3 \times 3$ squares has at least 4 colored cells, and to minimize the number of colored cells; we first consider the overlapping parts of these 4 $3 \times 3$ squares;
The cell at the 3rd row and 3rd column is counted 4 times,
The cells at the 1st row and 3rd column, 2nd row and 3rd column, 4th row and 3rd column, 5th row and 3rd column,
3rd row and 1st column, 3rd row and 2nd column, 3rd row and 2nd column, 3rd row and 2nd column are each counted 2 times;
$$
4 \times 4=16=1 \times 4+2 \times 6 \text {; }
$$

Therefore, at least $1+6=7$ cells need to be colored;
On the other hand, as shown in Figure 2, is a valid coloring method;
In conclusion, at least 7 cells need to be colored.