To determine if there exists an integer $n$ such that $n^2 + n + 1$ is divisible by $1955$, we need to check the divisibility of $n^2 + n + 1$ by the prime factors of $1955$.

First, factorize $1955$:
$$1955 = 5 \times 17 \times 23$$

We need to check if $n^2 + n + 1$ can be divisible by each of these prime factors.

1. Check divisibility by 5:
$$n^2 + n + 1 \equiv 0 \pmod{5}$$
We need to check the possible values of $n$ modulo 5:
$$\begin{aligned} n &\equiv 0 \pmod{5} \implies n^2 + n + 1 \equiv 0^2 + 0 + 1 \equiv 1 \pmod{5} \\ n &\equiv 1 \pmod{5} \implies n^2 + n + 1 \equiv 1^2 + 1 + 1 \equiv 3 \pmod{5} \\ n &\equiv 2 \pmod{5} \implies n^2 + n + 1 \equiv 2^2 + 2 + 1 \equiv 7 \equiv 2 \pmod{5} \\ n &\equiv 3 \pmod{5} \implies n^2 + n + 1 \equiv 3^2 + 3 + 1 \equiv 13 \equiv 3 \pmod{5} \\ n &\equiv 4 \pmod{5} \implies n^2 + n + 1 \equiv 4^2 + 4 + 1 \equiv 21 \equiv 1 \pmod{5} \end{aligned}$$
None of these results are congruent to 0 modulo 5. Therefore, $n^2 + n + 1$ is never divisible by 5.

Since $n^2 + n + 1$ is not divisible by 5, it cannot be divisible by $1955$ because 5 is a factor of 1955.

$\blacksquare$