2005 AMC 10B Problem 22

Use $1+2+\cdots+n=\dfrac{n(n+1)}{2}$ and simplify $\dfrac{n!}{\frac{n(n+1)}{2}}$

The reduced fraction $\dfrac{2(n-1)!}{n+1}$ fails to be an integer exactly when $n+1$ is an odd prime

Since $1 + 2 + \cdots + n = \dfrac{n(n+1)}{2},$ divisibility is equivalent to $$\dfrac{n!}{\frac{n(n+1)}{2}} = \dfrac{2(n-1)!}{n+1}$$ being an integer.

Put $N=n+1.$ If $N$ is composite and not a square, it has two distinct proper factors whose product is $N;$ both occur in $(N-2)!=(n-1)!.$ If $N=k^2$ with $k\ge3,$ the factors $k$ and $2k$ occur in that factorial, so it contains a multiple of $2N.$ The remaining composite case, $N=4,$ also divides $2(N-2)!=4.$ Thus the fraction is an integer whenever $N$ is composite. If $N=n+1$ is an odd prime, it divides neither $(n-1)!$ nor $2,$ so the fraction is not an integer. The even prime $N=2$ gives $n=1,$ which works.

The odd primes at most $25$ are $3,$ $5,$ $7,$ $11,$ $13,$ $17,$ $19,$ $23,$ giving $8$ failing values of $n.$ Hence $24 - 8 = 16$ values work.

Thus, C is the correct answer.