jmc

In symbols, we want the sum of the positive integers $k$ such that $$\frac{2}{3}<\frac{k}{27}<\frac{8}{9}.$$ Multiplying the expressions in these inequalities by $27$, we have $$18<k<24.$$ The integer solutions are thus $k=19,20,21,22,23$, which add up to $5\cdot 21=\boxed{105}$.