Because $2017$ is a prime number. All positive integers less than $2017$ are co-prime to it. Meanwhile, $\varphi(2017)=2016$. Therefore by Euler's theorem, we have $$N^{2016}\equiv 1\pmod{2017}\implies (N^{2016})^{2016}\equiv 1\pmod{2017}$$

This means all such $N$s satisfy the requirement. Hence, the answer is $\boxed{2016}$.