imc

Since $1+2+\cdots +n=\frac{n(n+1)}{2}$, the condition is equivalent to having an integer value for $\frac{n!}{\frac{n(n+1)}{2}}$. This reduces, when $n\ge 1$, to having an integer value for $\frac{2(n-1)!}{n+1}$. This fraction is an integer unless $n+1$ is an odd prime. There are $8$ odd primes less than or equal to $24$, so there are $24-8=\boxed{16}$ numbers less than or equal to $24$ that satisfy the condition.