jmc

Note that $27$ and $40$ are relatively prime, so $27$ has an inverse $(\mathrm{mod}40)$. Conveniently, the inverse of $27(\mathrm{mod}40)$ is easily found to be $3$, as we have $27\cdot 3=81\equiv 1(\mathrm{mod}40)$.

To solve the congruence $27a\equiv 17(\mathrm{mod}40)$, we multiply both sides by $3$ and simplify: $$\begin{array}{rl}3\cdot 27a& \equiv 3\cdot 17(\mathrm{mod}40)\\ a& \equiv 51(\mathrm{mod}40)\\ a& \equiv 11(\mathrm{mod}40)\end{array}$$Each operation in this sequence is reversible, so the solution set is exactly the set of integers congruent to $11(\mathrm{mod}40)$. The smallest and second-smallest positive solutions are $11$ and $51$. Their sum is $\boxed{62}$.