2. Given an $n \times n$ ($n$ is an odd number) chessboard where each unit square is colored in a checkerboard pattern, and the 4 corner unit squares are colored black. A figure formed by 3 connected unit squares in an L-shape is called a "domino". For what value of $n$ can all the black squares be covered by non-overlapping "dominoes"? If it can be covered, what is the minimum number of "dominoes" needed?
Solution: Let $n=2m+1$, consider the odd rows, then each row has $m+1$ black cells, with a total of $(m+1)^2$ black cells. Any two black cells cannot be covered by a single "domino", therefore, at least $(m+1)^2$ "dominoes" are needed to cover all the black cells on the chessboard. Since when $n=1,3,5$, we have $3(m+1)^2 > n^2$, thus, $n \geqslant 7$.

Below, we use mathematical induction to prove: when $n \geqslant 7$, $(m+1)^2$ "dominoes" can cover all the black cells on the chessboard.

When $n=7$, since two "dominoes" can form a $2 \times 3$ rectangle, and two $2 \times 3$ rectangles can form a $4 \times 3$ rectangle, then 4 such $4 \times 3$ rectangles can be placed on a $7 \times 7$ chessboard, such that all cells except the middle black cell are covered (as shown in Figure 1). Adjust the "domino" adjacent to the middle black cell, so that this "domino" covers the middle black cell and also covers the only black cell that the original "domino" covered. Thus, 16 "dominoes" cover all cells except one white cell on the chessboard (as shown in Figure 1).

Assume that when $n=2m-1$, on a $(2m-1) \times (2m-1)$ chessboard, $m^2$ "dominoes" can cover all the black cells. When $n=2m+1$, divide the $(2m+1) \times (2m+1)$ chessboard into three parts: $(2m-1) \times (2m-1)$, $(2m-1) \times 2$, and $(2m+1) \times 2$. Since the $(2m-1) \times 2$ rectangle can be divided into $m-2$ $2 \times 2$ squares and one $2 \times 3$ rectangle, the black cells in the $(2m-1) \times 2$ rectangle can be covered by $(m-2) + 2$ "dominoes". Similarly, the $(2m+1) \times 2$ rectangle can be covered by $(m-1) + 2$ "dominoes" (as shown in Figure 2). Therefore, the $(2m+1) \times (2m+1)$ chessboard can be covered by $m^2 + m + (m+1) = (m+1)^2$ "dominoes".