The [set](https://artofproblemsolving.com/wiki/index.php/Set) $\{3,6,9,10\}$ is augmented by a fifth element $n$, not equal to any of the other four. The [median](https://artofproblemsolving.com/wiki/index.php/Median) of the resulting set is equal to its [mean](https://artofproblemsolving.com/wiki/index.php/Mean). What is the sum of all possible values of $n$?
$\mathrm{(A)}\ 7\qquad \mathrm{(B)}\ 9\qquad \mathrm{(C)}\ 19\qquad \mathrm{(D)}\ 24\qquad \mathrm{(E)}\ 26$
The median must either be $6, 9,$ or $n$. [Casework](https://artofproblemsolving.com/wiki/index.php/Casework):

Median is $6$: Then $n \le 6$ and $\frac{3+6+9+10+n}{5} = 6 \Longrightarrow n = 2$.
Median is $9$: Then $n \ge 9$ and $\frac{3+6+9+10+n}{5} = 9 \Longrightarrow n = 17$.
Median is $n$: Then $6 < n < 9$ and $\frac{3+6+9+10+n}{5} = n \Longrightarrow n = 7$.
All three cases are valid, so our solution is $2 + 7 + 17 = 26 \Longrightarrow \mathrm{(E)}$.