3. Color some nodes (vertices of unit squares) of a $6 \times 6$ grid table red, such that the boundary of any sub-grid table $k \times k (1 \leqslant k \leqslant 6)$ formed by unit squares has at least one red point. Find the minimum number of red points that satisfy this condition.
3. The minimum value is 12.

On one hand, represent the nodes as coordinates $(i, j) (i, j \in \{-3, -2, -1, 0, 1, 2, 3\})$, and color the following 12 points $(\pm 3, 0), (0, \pm 3), (\pm 1, \pm 1), (\pm 2, \pm 2)$ red. By checking each one, it is known that they satisfy the condition.

On the other hand, use proof by contradiction to show that if there are only 11 red points, there exists a sub-square grid whose boundary has no red points.

Assume that a coloring with exactly 11 red points satisfies the condition.

Notice that, in a $2 \times 2$ sub-square grid, at least two nodes are red.

As shown in Figure 7, the $2 \times 2$ sub-square grid formed by four C's has at least two red points, each A has at least one red point, and each pair of adjacent B's has at least one red point. Thus, there are already at least 10 red points. Therefore, the total number of red points in the four pairs of adjacent B's is 4 or 5.
(1) If it is 4, then each pair of adjacent B's has exactly one red point. Clearly, these red points can only appear at their common nodes. At this point, consider the $2 \times 2$ sub-square grids at the four corners containing A, each of which has at least two red points. Thus, the number of red points is at least $4 + 4 \times 2 = 12$, which contradicts the assumption.
(2) If it is 5, then exactly one pair of adjacent B's has 2 red points, and the other three pairs of adjacent B's each have 1 red point. Clearly, these red points can only appear at their common nodes. Consider the $2 \times 2$ sub-square grids at the two lower corners containing A, each of which has at least two red points, and these red points do not appear in the two upper C grids, which also have at least one red point each. Thus, the total number of red points is at least $3 \times 2 + 6 \times 1 = 12$, which contradicts the assumption.