[Example 5.6.4] On a $6 \times 6$ chessboard, $n$ dominoes of size $1 \times 2$ are placed, each domino covering exactly two squares. If no matter how these $n$ dominoes are placed, it is always possible to place one more domino, find the maximum value of $n$.


Obviously, placing 12 dominoes as shown in the figure leaves no room for another domino, so $n \leqslant 11$.

We prove that 11 dominoes can meet the requirements of the problem.
Consider the following two sets:
$A=\{$ the empty cells in the lower $5 \times 6$ grid $\}$,
$B=\{$ the dominoes in the upper $5 \times 6$ grid $\}$.
If for 11 dominoes, no more dominoes can be placed, there must be a correspondence $f$ such that:
(1) Each empty cell in $A$ must have a domino above it, otherwise, if the cell above is still empty, due to the definition of $A$, two empty cells will inevitably be adjacent, allowing another domino to be placed.
(2) Different empty cells in $A$ must correspond to different dominoes, otherwise, if different cells correspond to the same domino, these two different cells will inevitably be adjacent, allowing another domino to be placed.
Thus, $f$ is a mapping from $A$ to $B$.
Since the number of dominoes is no more than 11, then
$$
|A| \leqslant|B| \leqslant 11 \text {. }
$$

Moreover, the top row can have at most 3 empty cells, otherwise, if there are 4 empty cells, two of them will inevitably be adjacent, allowing another domino to be placed.

Since there are a total of $36-22=14$ empty cells, $A$ must have at least 11 empty cells, i.e., $|A| \geqslant 11$.
Thus,
$$
|A|=|B|=11 .
$$

By the definition of $B$, there are 11 dominoes in the upper $5 \times 6$ grid, so the last row must be empty, allowing another domino to be placed: