First note that $n \equiv 1 \pmod{19}$ implies $n^5 + n^4 + n^3 + n^2 + n + 1 \equiv 6 \pmod{19}$. Assume now that $n \not\equiv 1 \pmod{19}$, 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
$$19 \mid n^5 + n^4 + n^3 + n^2 + n + 1 \iff n^6 \equiv 1 \pmod{19}.$$
Because 19 is a prime number, the congruence $a^2 \equiv 1 \pmod{19}$ has exactly two solutions, namely $a \equiv \pm 1 \pmod{19}$. 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 $n^6 \equiv 1 \pmod{19}$ iff $n^3 \equiv \pm 1 \pmod{19}$.
To find all such $n$ we create the following table, in which we first calculated $n^2 \pmod{19}$ 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).