jmc

The only possibilities for the numbers are $11,12,13,14,15,16$ and $10,11,12,13,14,15$. In the second case, the common sum would be $(10+11+12+13+14+15)/3=25$, so $11$ must be paired with $14$, which it isn't. Thus, the only possibility is the first case and the sum of the six numbers is $81\to \boxed{81}$.