jmc

Since $\overline{ab}|\overline{abcd}=100\cdot \overline{ab}+\overline{cd}$, then $\overline{ab}$ also divides into $\overline{abcd}-100\cdot \overline{ab}=\overline{cd}$. Similarly, since $\overline{cd}|\overline{abcd}=100\cdot \overline{ab}+\overline{cd}$, then $\overline{cd}$ must divide into $\overline{abcd}-\overline{cd}=100\cdot \overline{ab}$. To minimize $\overline{abcd}$, then we would like to try $a=b=1$. It follows that $\overline{cd}$ is divisible by $11$, and also divides into $100\cdot \overline{ab}=1100$. Thus, $\overline{cd}=11,22,44,55$, but we can eliminate the first due to the distinctness condition. Trying each of the others, we see that $1122=2\cdot 3\cdot 11\cdot 17$ is not divisible by $12$; $1144=2^3\cdot 11\cdot 13$ is not divisible by $14$; and $\boxed{1155}=3\cdot 5\cdot 7\cdot 11$ is indeed divisible by $15$.