8. In an $8 \times 8$ grid, what is the minimum number of small squares that need to be removed so that it is impossible to cut out a complete "T-shaped" pentomino from the remaining grid?
（Sun Wenxian, problem contributor）

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8. At least 14 small squares must be removed from Figure 6.
Figure 6
Figure 7

As shown in Figure 7, an $8 \times 8$ chessboard is divided into five regions. The central region must have at least two small squares removed to prevent a T-shaped pentomino from being placed. The two crossed positions are not equivalent; one is in a corner and the other is in an internal position. Removing just one of them is insufficient to prevent the T-shaped pentomino from being placed.

The following proof shows that for the four congruent regions on the boundary, each region must have at least 3 small squares removed to prevent a T-shaped pentomino from being placed.

As shown in Figure 8, using the upper right region as an example, the T-shaped part below must have 1 small square removed, and the part above must have 1 small square removed at the crossed position.
The T-shaped part below must have 1
Figure 8 small square removed, with five possible cases. If the small square at the crossed position in Figure 9 is removed, another T-shaped pentomino can still be placed; the same applies to the other cases (as shown in Figure 10). Therefore, at least 3 small squares must be removed.
Figure 9
Figure 10
In summary, for the T-shaped pentomino, at least $3 \times 4 + 2 = 14$ small squares must be removed.
(Provided by Tao Pingsheng)