Note that $(n-1)(n^5 + n^4 + n^3 + n^2 + n + 1) = n^6 - 1 = (n^3 - 1)(n^3 + 1) = (n-1)(n+1)(n^2 - n + 1)(n^2 + n + 1)$, and so
$$n^5 + n^4 + n^3 + n^2 + n + 1 = (n+1)(n^2 - n + 1)(n^2 + n + 1).$$
Because 199 is a prime number, this expression is divisible by 199 iff one of the factors $n+1$, $n^2-n+1$ or $n^2+n+1$ is divisible by 199. Divisibility of $n+1$ by 199 is equivalent to $n \equiv 198 \pmod{199}$.
To solve the quadratic congruence $n^2 - n + 1 \equiv 0 \pmod{199}$, we observe $n^2 - n + 1 \equiv n^2 - 200n + 1 \equiv (n - 100)^2 - 100^2 + 1 \pmod{199}$. Because $100^2 - 1 = 200 \cdot 50 - 1 = 199 \cdot 50 + 49 \equiv 49 \pmod{199}$, the congruence $n^2 - n + 1 \equiv 0 \pmod{199}$ is equivalent to $(n - 100)^2 \equiv 49 \pmod{199}$. Again, because 199 is a prime, this has exactly two solutions (mod 199) which are determined by $n - 100 \equiv \pm 7 \pmod{199}$. These two solutions are $n \equiv 107 \pmod{199}$ and $n \equiv 93 \pmod{199}$.
Note now that that $(n-1)^2+(n-1)+1 = n^2-n+1$, hence $n \equiv 106 \pmod{199}$ and $n \equiv 92 \pmod{199}$ are the solutions of $n^2+n+1 \equiv 0 \pmod{199}$.
This shows that $n^5+n^4+n^3+n^2+n+1$ is divisible by 199 iff $n$ is congruent to 92, 93, 106, 107 or 198 (mod 199).