any ideas?

By the third condition, it is clear that all digits of $N$ are even. Also notice that if the ten's and unit's digit in a number are even and the number is divisible by $4$, then the unit's digit is also divisible by $4$. Hence, again by condition 3, we get that all the digits of $N$ are divisible by $4$, which means they are all $4$ or $8$. Say there are $a$ $4$'s and $(10 - a)$ $8$'s. Then the digit sum of the number is $4a + 8(10 - a) = 80 - 4a \equiv 2a - 1 (\text{mod 3})$. Therefore $a \equiv 2 ( \text{mod 3}) \implies a = 2, 5, 8$. Now follows some casework using $1000 \equiv -1 (\text{mod 11})$ which I am too lazy to do here .