Four, for a $7 \times 7$ grid of small squares, how many of the small squares can be shaded such that no two shaded squares are adjacent? Adjacent means they share a side. 

How many of the small squares can be shaded?
Four, color at most 26 small squares.
The coloring in Figure 2 satisfies the conditions.

The following proves: At most 26 small squares can be colored.

First, according to the problem, for any $2 \times 2$ square grid, at most two of the small squares can be colored; for a $3 \times 3$ square grid, at most 5 of the small squares can be colored to meet the conditions.

Second, for a $5 \times 5$ square grid, it is based on a $3 \times 3$ square grid with a border of width 2 added (as shown in Figure 3). This border can be divided into 4 $2 \times 2$ square grids, with the bottom-right two squares, one being reused and the other not used. Therefore, at most 9 additional squares can be colored, and when adding 9 colored squares, the bottom-right square must be colored, while the reused square is not colored. Thus, a $5 \times 5$ square grid can have at most 14 squares colored, and when 14 squares are colored, the bottom-right square is colored.

Finally, a $7 \times 7$ square grid is based on a $5 \times 5$ square grid with a border of width 2 added (as shown in Figure 4).

Similarly, at most 13 additional squares can be colored, and when adding 13 colored squares, the bottom-right square must be colored, while the reused square is not colored. Thus, at most 27 squares can be colored. At this point, assume the square in the last row and the second-to-last column is not colored (otherwise, the square in the second-to-last row and the last column is not colored). According to the previous analysis, the two squares in the last two rows and the third-to-last column are colored, which is a contradiction.
Therefore, at most 26 squares can be colored.
In summary, at most 26 small squares can be colored.