Ireland

First note that $n\equiv 1(\mathrm{mod}19)$ implies $n^5+n^4+n^3+n^2+n+1\equiv 6(\mathrm{mod}19)$. Assume now that $n\not\equiv 1(\mathrm{mod}19)$, i.e. $\gcd (19,n-1)=1$. Because $n^6-1=(n-1)(n^5+n^4+n^3+n^2+n+1)$ we then have $$19\mid n^5+n^4+n^3+n^2+n+1⟺n^6\equiv 1(\mathrm{mod}19).$$ Because 19 is a prime number, the congruence $a^2\equiv 1(\mathrm{mod}19)$ has exactly two solutions, namely $a\equiv \pm 1(\mathrm{mod}19)$. This is so because $(a-1)(a+1)$ can only be divisible by the prime number 19 if one of the two factors is so. This shows that $n^6\equiv 1(\mathrm{mod}19)$ iff $n^3\equiv \pm 1(\mathrm{mod}19)$. To find all such $n$ we create the following table, in which we first calculated $n^2(\mathrm{mod}19)$ to keep the numbers small.

n (mod 19)	0	-1	±2	±3	±4	±5	±6	±7	±8	±9
$n^2$ (mod 19)	0	1	4	9	-3	6	-2	-8	7	5
$n^3$ (mod 19)	0	-1	±8	±8	±7	∓8	±7	±1	∓1	±7

This shows that $n^5+n^4+n^3+n^2+n+1$ is divisible by 19 if and only if $n$ is congruent to 7, 8, 11, 12 or 18 (mod 19).