How many different positive integers divisible by 4 can be formed using each of the digits 1, 2, 3, and 4 at most once, and no other digits? For example, 12 counts, but 512 does not.
The only one-digit integer divisible by $4$ that we can construct is $4$.

We can construct $3$ two-digit integers divisible by $4$: $12$, $32$, and $24$.

An integer is divisible by $4$ if its rightmost two digits are divisible by $4$.  Thus we can append either or both of the remaining two digits to any of these two-digit integers and preserve divisibility by $4$.  For each, there are $2$ ways to choose one digit to append, and $2$ ways to order the digits if we append both of them.  Thus we get $4$ more integers for each, or $12$ total.

The full number is $12+3+1=\boxed{16}$ integers.