Example 12 (1990 Balkan Mathematical Olympiad) Let the sequence $\left\{a_{n}\right\}$ satisfy $a_{1}=1, a_{2}=3$, and for all $n \in \mathbf{N}$, $a_{n+2}=(n+3) a_{n+1}-(n+2) a_{n}$. Find all values of $n$ for which $a_{n}$ is divisible by 11.
Given $a_{n+2}-a_{n+1}=(n+2)\left(a_{n+1}-a_{n}\right)$, we construct a new sequence $\left\{b_{n}\right\}(n \geqslant 2), b_{n+1}=a_{n+1}-a_{n}(n \geqslant 1)$, then $b_{2}=2, b_{n+1}=(n+1)\left(a_{n}-a_{n-1}\right)=(n+1) b_{n}(n \geqslant 2)$,
$$
\therefore b_{n}=n b_{n-1}=n(n-1) b_{n-2}=\cdots=n(n-
$$
1) $\cdots 3 b_{2}=n! \quad(n \geqslant 2)$,
$$
\begin{array}{l}
\quad \therefore a_{n}=a_{1}+\sum_{k=2}^{n}\left(a_{n}-a_{n-1}\right)=1+\sum_{k=2}^{n} b_{k} \\
=\sum_{k=1}^{n} k!
\end{array}
$$

Thus, $a_{4}=11 \times 3, a_{8}=11 \times 4203, a_{10}=11 \times 367083$, for $n \geqslant 11$, since $\sum_{k=1}^{10} k!$ is divisible by 11, therefore $a_{n}=\sum_{k=1}^{10} k!+\sum_{k=11}^{n} k!$ is also divisible by 11. Hence, the values of $n$ are $n=4,8$ and all natural numbers $n \geqslant 10$.