Question 1209684
Let the given expression be equal to an integer $k$.
\[ \frac{n^2 + n + 1}{n^3 + n - 2} = k \]
\[ n^2 + n + 1 = k(n^3 + n - 2) \]
\[ n^2 + n + 1 = kn^3 + kn - 2k \]
\[ kn^3 - n^2 + (k-1)n - (1+2k) = 0 \]

If $k=0$, then $n^2 + n + 1 = 0$, which has no integer solutions.

If $n=0$, then $\frac{1}{-2} = k$, which is not an integer.
If $n=1$, then $\frac{3}{0}$, which is undefined.
If $n=2$, then $\frac{7}{8}$, which is not an integer.
If $n=3$, then $\frac{13}{28}$, which is not an integer.

If $n$ is large, then $n^3$ dominates, so $k$ must be close to 0.

If $k=1$, we have $n^3 - n^2 + 0n - 3 = 0$.
$n^3 - n^2 - 3 = 0$.
If $n=2$, $8-4-3 = 1 \ne 0$.
If $n=3$, $27-9-3 = 15 \ne 0$.

If $n^3 + n - 2 = 1$, then $n^3 + n - 3 = 0$.
If $n=1$, $1+1-3 = -1$.
If $n=2$, $8+2-3 = 7$.

If $n^3 + n - 2 = -1$, then $n^3 + n - 1 = 0$.
If $n=0$, $-1$.
If $n=1$, $1$.

If $n=2$, $n^3+n-2 = 8+2-2 = 8$. Then $\frac{4+2+1}{8} = \frac{7}{8}$.

Consider the case where $n$ is large. Then $k \approx \frac{n^2}{n^3} = \frac{1}{n}$.
If $n$ is large, $k$ must be close to 0.

If $k=0$, $n^2+n+1=0$, no integer solution.

If $n=1$, $\frac{3}{0}$ undefined.
If $n=2$, $\frac{7}{8}$.
If $n=3$, $\frac{13}{28}$.

If $n=-1$, $\frac{1-1+1}{-1-1-2} = \frac{1}{-4}$.
If $n=-2$, $\frac{4-2+1}{-8-2-2} = \frac{3}{-12} = -\frac{1}{4}$.
If $n=-3$, $\frac{9-3+1}{-27-3-2} = \frac{7}{-32}$.

Final Answer: The final answer is $\boxed{no solutions}$