Claim: All $n$ of the form $18k-9 $ ($k$ is a positive integer) satisfy the given condition Proof: putting $n = 18k-9, $ we see that $(2^9)^{2k-1} \equiv {-1} \pmod{19}$ $(3^9)^{2k-1} \equiv {-1} \pmod{19}$ $(4^9)^{2k-1} \equiv {1} \pmod {19}$ Hence, $1+2^{18k-9}+3^{18k-9}+4^{18k-9} \equiv{0} \pmod{19}$