9. Let $a_{1}=1, a_{2}=3$, and for all positive integers $n$, $a_{n+2}=(n+3) a_{n+1}-$ $(n+2) a_{n}$, find all values of $n$ such that $a_{n}$ is divisible by 11.
9. Let $b_{n+1}=a_{n+1}-a_{n}, n=1,2,3, \cdots$, then from $a_{n+2}=(n+3) a_{n+1}-(n+2) a_{n}$, we get $b_{n+1}=(n+1) b_{n},(n=2,3, \cdots)$.

Since $b_{2}=2=2!$, then $b_{n}=n!(n=2,3, \cdots)$, so $a_{n}=\sum_{i=1}^{n} i!$, and after calculation, we find that when $n=4 ; 8$, $a_{n}$ is divisible by 11.

When $n \geqslant 10$, $a_{n}=(1!+2!+\cdots+8!)+(9!+10!)+(11!+\cdots+n!)$, and the sum of each bracket can be divided by 11, so when $n \geqslant 10$, $a_{n}$ is also divisible by 11.
Therefore, the $n$ that satisfies the condition is $n=4,8$ or $n \geqslant 10$.