I claim the answer is all odd $n$. We have two cases based on the parity of $n$. Case 1: n is odd. Say $n=2m+1$ for some positive integer $m$. Then, the solution $(a,b)=(27\times 2018^m, 13\times 2018^m)$ exists and works. Case 2: n is even. We take $\pmod{3}$ to see that $-b^2\equiv 1\pmod{3}$, which has no integer solutions in $b$.