For how many positive integers $n$ less than or equal to $24$ is $n!$ evenly divisible by $1 + 2 + \cdots + n?$
$\textbf{(A) } 8 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 21$
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 $24 - 8 =  \boxed{\textbf{(C) }16}$ numbers less than or equal to $24$ that satisfy the condition.