jmc

Multiplying both sides of the equation by $3jk$ to clear the denominator gives $3k+3j=jk$. Re-arranging and applying Simon's Favorite Factoring Trick, it follows that $$jk-3j-3k+9=(j-3)(k-3)=9.$$ Thus, $j-3$ and $k-3$ are pairs of positive factors of $9$, so $(j-3,k-3)=(1,9),(3,3),(9,1)$. These give $k=4,6,12$, and their sum is $4+6+12=\boxed{22}$.