Find the number of positive integers with three not necessarily distinct digits, $abc$, with $a \neq 0$ and $c \neq 0$ such that both $abc$ and $cba$ are multiples of $4$.
Whether or not an integer is divisible by $4$ is determined by its last two digits. Because $a$ and $c$ cannot be zero, therefore we need to perform casework.
This means that regardless of the parity of $b$, there are always two choices each for $a$ and $c$. It follows that because $b$ has $10$ choices, hence the final answer is $10\times 2\times 2=\boxed{40}$.