For how many positive integers $n$ does $1+2+...+n$ evenly divide from $6n$? 
$\textbf{(A) } 3\qquad \textbf{(B) } 5\qquad \textbf{(C) } 7\qquad \textbf{(D) } 9\qquad \textbf{(E) } 11$
If $1+2+...+n$ evenly [divides](https://artofproblemsolving.com/wiki/index.php/Divide) $6n$, then $\frac{6n}{1+2+...+n}$ is an [integer](https://artofproblemsolving.com/wiki/index.php/Integer). 
Since $1+2+...+n = \frac{n(n+1)}{2}$ we may substitute the [RHS](https://artofproblemsolving.com/wiki/index.php?title=RHS&action=edit&redlink=1) in the above [fraction](https://artofproblemsolving.com/wiki/index.php/Fraction). So the problem asks us for how many [positive integers](https://artofproblemsolving.com/wiki/index.php/Positive_integer) $n$ is $\frac{6n}{\frac{n(n+1)}{2}}=\frac{12}{n+1}$ an integer, or equivalently when $k(n+1) = 12$ for a positive integer $k$. 
$\frac{12}{n+1}$ is an integer when $n+1$ is a [factor](https://artofproblemsolving.com/wiki/index.php/Divisor) of $12$. 
The factors of $12$ are $1, 2, 3, 4, 6,$ and $12$, so the possible values of $n$ are $0, 1, 2, 3, 5,$ and $11$.
But since $0$ isn't a positive integer, only $1, 2, 3, 5,$ and $11$ are the possible values of $n$. Therefore the number of possible values of $n$ is $\boxed{\textbf{(B) }5}$.