2019 AMC 10A Problem 9

The sum $1+2+\cdots+n$ equals $\frac{n(n+1)}{2}$

The obstruction comes from when $n+1$ contributes a new prime factor

The sum of the first $n$ numbers is $$\dfrac{n(n + 1)}{2}.$$ We need this to not divide $n!.$

Put $m=n+1$. If $m$ is composite, write $m=ab$ with $2\le a\le b$. When $a<b$, the distinct factors $a$ and $b$ both occur in $(m-2)!=(n-1)!$. When $a=b$, we have $a\ge3$, and the two multiples $a$ and $2a$ both occur in $(m-2)!$, so $a^2=m$ divides that factorial as well. Thus $(n-1)!$ is divisible by $m$, and consequently $$n! \text{ is divisible by } \frac{n(n+1)}2.$$

Conversely, if $n+1$ is prime, that prime factor does not occur in $n!$, so the divisibility fails. Since $997$ is prime while $998,999,$ and $1000$ are composite, the greatest three-digit value is $$997-1=996.$$

Thus, B is the correct answer.