8. A square of size $3 \times 3$ with a unit square removed from each corner is called a "cross shape". On a $10 \times 11$ chessboard, what is the maximum number of non-overlapping "cross shapes" (each "cross shape" exactly covers 5 small squares on the chessboard)?
(Supplied by Feng Zuming)
8. 15.

First, prove that the maximum number of "crosses" that can be placed is 15.
Proof by contradiction. Assume that 16 "crosses" can be placed.
For each "cross", we call the central square the "center" (denoted as *). As shown in Figure 3, remove the outermost layer of squares from the $10 \times 11$ chessboard, resulting in an $8 \times 9$ grid. Clearly, the "center" of each "cross" can only appear within this $8 \times 9$ grid.
Note that in every $3 \times 3$ grid, at most two "centers" can be placed, as shown in Figure 4.
Next, discuss how to place as many "centers" as possible in an $8 \times 3$ grid. As shown in Figure 5, divide the $8 \times 3$ grid into three sub-grids: $3 \times 3$, $2 \times 3$, and $3 \times 3$. If at most one "center" is placed in the middle $2 \times 3$ grid, since at most two "centers" can be placed in each $3 \times 3$ grid, a maximum of 5 "centers" can be placed in total. If two "centers" are placed in the middle $2 \times 3$ grid, then the third and sixth rows must be empty. Through exhaustive enumeration, it is known that there are only two ways to place 6 "centers" as shown in Figure 5.

As shown in Figure 6, divide the $8 \times 9$ grid into three $8 \times 3$ grids, labeled (a), (b), and (c) from left to right. Since there are 16 "centers" in the $8 \times 9$ grid, one of the $8 \times 3$ grids must contain 6 "centers".

If (b) contains 6 "centers", since there are only two symmetric ways to place 6 "centers" in an $8 \times 3$ grid, assume the 6 "centers" are distributed as shown in Figure 6. It is easy to see that the 3rd and 7th columns cannot contain "centers". It is not difficult to see that (a) and (c) can each contain at most 4 "centers". Thus, a total of $4 + 6 + 4 = 14$ "centers" are placed, leading to a contradiction.

If (a) and (c) each contain 6 "centers", then the 4th and 6th columns cannot contain "centers", so (b) can only place "centers" in the middle column. It is not difficult to see that (b) can contain at most 3 "centers". Thus, a total of $6 + 3 + 6 = 15$ "centers" are placed, also leading to a contradiction.

If (a) or (c) contains 6 "centers", assume (a) contains 6 "centers" and they are placed as shown in Figure 7. In this case, the 4th column cannot contain "centers", so (b) can contain at most 4 "centers". Thus, a total of $6 + 4 + 5 = 15$ "centers" are placed, also leading to a contradiction.

In summary, it is known that at most 15 "centers" can be placed in the $8 \times 9$ grid.

Finally, provide an example showing that 15 "centers" can indeed be placed, as shown in Figure 8.