24.3.9 * A shape obtained by removing a unit square from each corner of a $3 \times 3$ square is called a “cross-shape”.

On a $10 \times 11$ chessboard, what is the maximum number of non-overlapping “cross-shapes” that can be placed? (Each “cross-shape” exactly covers 5 small squares on the chessboard.)
Analysis Answer: 15.

First, prove that the maximum number of "crosses" that can be placed is 15. Use proof by contradiction, assuming that 16 "crosses" can be placed.

For each "cross," we call the central square the "heart" (denoted as *). Remove the outermost layer of squares from the $10 \times 11$ chessboard, resulting in an $8 \times 9$ grid. Clearly, the "heart" of each "cross" can only appear within this $8 \times 9$ grid.
Note that in every $3 \times 3$ grid, at most two "hearts" can be placed:
We will discuss how to place as many "hearts" as possible in an $8 \times 3$ grid. Divide the $8 \times 3$ grid from top to bottom into three sub-grids: $3 \times 3$, $2 \times 3$, and $3 \times 3$. If one "heart" is placed in the middle $2 \times 3$ grid, then since at most two "hearts" can be placed in each $3 \times 3$ grid, a total of 5 "hearts" can be placed. If two "hearts" are placed in the middle $2 \times 3$ grid, then a total of 6 "hearts" can be placed in the $8 \times 3$ grid, at which point the third and sixth rows must be empty. Through exhaustive enumeration, it is known that there are only the two placement methods shown in the following diagram:
Now, divide the $8 \times 9$ grid into three $8 \times 3$ grids, from left to right, called grid (a), grid (b), and grid (c). Since there are 16 "hearts" in the $8 \times 9$ grid, one of the $8 \times 3$ grids must have 6 "hearts".

If grid (b) has 6 "hearts," since there are only two mutually symmetrical ways to place 6 "hearts" in an $8 \times 3$ grid, we can assume the 6 "hearts" are distributed as shown in the left diagram below:

It is easy to see that the 3rd and 7th columns cannot have "hearts." Thus, it is not difficult to see that at most 4 "hearts" can be placed in grids (a) and (c), resulting in a total of $4+6+4=14$ "hearts," leading to a contradiction.

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

If 6 "hearts" are placed in either grid (a) or grid (c), assume 6 "hearts" are placed in grid (a), and they are placed as shown in the right diagram above. In this case, the 4th column cannot have "hearts," so at most 4 "hearts" can be placed in grid (b), resulting in a total of $6+4+5=15$ "hearts," also leading to a contradiction.
In summary, it is known that at most 15 "hearts" can be placed in the $8 \times 9$ grid.
Finally, in addition to the two examples already provided, we give two more examples to show that indeed 15 "hearts" can be placed: