Since the RHS is always divisible by $2$ we only have to check for $1009$.By fermat if $k$ works then so does $k+1008$.Hence finding one such number suffices.Found some $n=1009k+252,1009k+253,1009k+756,1009k+757$ all work.@below what do have to say now

Pluto1708 wrote: Since the RHS is always divisible by $2$ we only have to check for $1009$.By fermat if $k$ works then so does $k+1008$.Hence finding one such number suffices.Thoygh I haven't found even one till now. lol everyone knows that

$n = 1009k + 252$ works for every nonnegative integer $k.$ Note that $2^{252} \equiv -1 \pmod{1009}, 3^{252} \equiv -1 \pmod{1009},$ and $4^{252} \equiv 1 \pmod{1009}$ Then $1009 | 1 + 2^{252} + 3^{252} + 4^{252}$ and it is easy to see that $1 + 2^n + 3^n + 4^n$ is even for evert positive integer $n,$ and so we are done.

@above 252 can also be replaced by 253,756,757

Pluto1708 wrote: @above 252 can also be replaced by 253,756,757 Yes. We have: $$2018 | (1 + 2^n + 3^n + 4^n) \text{ if and only if } n \equiv 252, 253, 756, 757, 1260, 1261, 1764, 1765 \ (\text{mod } 2018)$$