(12) Let $n \geqslant 11$ be a positive integer, and let set $A$ consist of the sums of 10 consecutive positive integers not exceeding $n$, and set $B$ consist of the sums of 11 consecutive positive integers not exceeding $n$. If the number of elements in $A \cap B$ is 181, find the maximum and minimum values of $n$.
12 Clearly,
$$
\begin{array}{l}
A=\{55+10 k \mid 0 \leqslant k \leqslant n-10, k \in \mathbf{Z}\}, \\
B=\{66+11 l \mid 0 \leqslant l \leqslant n-11, l \in \mathbf{Z}\},
\end{array}
$$

To find the number of elements in $A \cap B$, let $55+10 k=66+11 l$, then $10 k=(l+1) 11$.
Let $k=11 m$, then we get $l=10 m-1$. Since $0 \leqslant k \leqslant n-10$, $m$ can take values $0,1,2, \cdots,\left[\frac{n-10}{11}\right]$, at this time the corresponding values of $l$ are $-1,9,19, \cdots$, $10\left[\frac{n-10}{11}\right]-1$. Note that
$$
10\left[\frac{n-10}{11}\right]-1 \leqslant 10 \cdot \frac{n-10}{11}-1 \leqslant n-11,
$$

which fits the range of $l$, discarding the value -1, then the number of elements in the set $A \cap B$ is $\left[\frac{n-10}{11}\right]$. Let $181=\left[\frac{n-10}{11}\right]$, then
$$
181 \leqslant \frac{n-10}{11}<182,
$$

which means $2001 \leqslant n<2012$, so the maximum and minimum values of $n$ are 2011 and 2001, respectively.