9. The answer is $(\mathbf{C})$. Consider the equations

$$
\left\{\begin{array}{l}
a_{n}=n\left(a_{1}+a_{2}+\cdots+a_{n-1}\right) \\
a_{n+1}=(n+1)\left(a_{1}+a_{2}+\cdots+a_{n-1}+a_{n}\right)
\end{array}\right.
$$

valid for every $n \geq 2$. We can rewrite the second equation as

$$
\begin{aligned}
a_{n+1} & =n\left(a_{1}+a_{2}+\cdots+a_{n-1}\right)+\left(a_{1}+a_{2}+\cdots+a_{n-1}\right)+(n+1) a_{n} \\
& =a_{n}+\frac{a_{n}}{n}+(n+1) a_{n} \\
& =\frac{n^{2}+2 n+1}{n} a_{n} \\
& =\frac{(n+1)^{2}}{n} a_{n}
\end{aligned}
$$

where, using the first equation, we have substituted $a_{1}+a_{2}+\cdots+a_{n-1}$ with $\frac{a_{n}}{n}$ wherever possible. We then obtain

$$
a_{n}=\frac{n^{2}}{n-1} a_{n-1}=\frac{n^{2}}{n-1} \frac{(n-1)^{2}}{n-2} a_{n-2}=\cdots=\frac{n^{2}}{n-1} \frac{(n-1)^{2}}{n-2} \cdots \frac{3^{2}}{2} a_{2} .
$$

Simplifying the denominator of each fraction with the numerator of the next and noting that $a_{2}=2$, we get

$$
a_{n}=n^{2} \cdot(n-1) \cdot(n-2) \cdot \ldots \cdot 4 \cdot \frac{3}{2} a_{2}=n \cdot \frac{n !}{2}
$$

for every $n \geq 2$. At this point, observe that $2022=2 \cdot 3 \cdot 337$ divides $n \cdot \frac{n !}{2}$ if and only if this number is divisible separately by 2, 3, and 337. On one hand, since $n \cdot \frac{n !}{2}$ is a product of numbers less than or equal to $n$ and 337 is prime, this product can be divisible by 337 only if $n \geq 337$. On the other hand, if $n=337$ we have $a_{n}=337 \cdot \frac{1}{2} \cdot 337!$, and $\frac{1}{2} \cdot 337! = 3 \cdot 4 \cdot \ldots \cdot 337$ is certainly a multiple of 6, so $2 \cdot 3 \cdot 337$ divides $a_{337}$, and thus 337 is the minimum sought.
9. Let $a_{1}=1$ and, for each $n \geq 2$,

$$
a_{n}=n\left(a_{1}+a_{2}+\cdots+a_{n-1}\right) .
$$

What is the smallest value of $n$ for which $a_{n}$ is divisible by 2022?
(A) 47
(B) 289
(C) 337
(D) 2022
(E) 2023