9. If a three-digit positive integer $n$ makes the sum of the first $n$ positive integers not a factor of the product of the first $n$ positive integers. Then the maximum value of $n$ is ( ).
(A) 995
( B ) 996
( C) 997
(D) 998
(E) 999
9. B.

Notice that,
$$
\frac{n!}{1+2+\cdots+n}=\frac{n!}{\frac{n(n+1)}{2}}=\frac{2((n-1)!)}{n+1} \text {. }
$$

If $n+1$ is composite, then $\frac{2((n-1)!)}{n+1} \in \mathbf{Z}$, which does not meet the requirement.

Therefore, $n+1$ must be a prime number. The largest three-digit prime number is 997.
Thus, the largest three-digit positive integer $n$ is 996.