A interesting combinatorics problem for beginners.

Let $\overline{ABCD}$ be such a number. Since it's a multiple of four, and is made up of distinct digits less than $5$ , the digits $\overline{CD}$ are one of $\{04,12,20,24,32,40\}$ .

If these two digits are both non-zero, then there are $2$ choices for $A$ (it can't be zero), and then $2$ choices for $B$ , for a total of $4$ choices.

If one of the digits $\overline{CD}$ is zero, then there are $3$ choices for $A$ , and $2$ for $B$ , for a total of $6$ choices.

In the list of possible $\overline{CD}$ pairs, there are three of each type; so the answer is $3\times 4+3\times 6=\boxed{30}$ .