jmc

Observing that $4\cdot 7=28=27+1,$ we multiply both sides of the given congruence by 7 to find $28x\equiv 91(\mathrm{mod}27)$. Since $28x\equiv x(\mathrm{mod}27)$ and $91\equiv 10(\mathrm{mod}27)$, we conclude that $x\equiv 10(\mathrm{mod}27)$. Therefore, $\boxed{10}$ is the smallest positive integer satisfying the given congruence.