1.97 Find the largest integer $n$ such that $\frac{(n-2)^{2}(n+1)}{2 n-1}$ is an integer.
(17th American Invitational Mathematics Examination, 1999)
[Solution] Let $k$ be a prime divisor of $2n-1$. That is, $k \mid 2n-1$.
Therefore, $k$ must be a divisor of $n-2$ or $n+1$,
i.e., $k \mid n-2$ or $k \mid n+1$.
If $k \mid n-2$, then $k \mid (2n-1) - (n-2) = n+1$,
If $k \mid n+1$, then $k \mid (2n-1) - (n+1) = n-2$.
Thus, $k \mid n-2$ and $k \mid n+1$.
Therefore, $k \mid (n+1) - (n-2)$, i.e., $k \mid 3$,
So $2n-1$ is a power of 3.
Let $2n-1 = 3^t$, then $n = \frac{3^t + 1}{2}$.
We have $\frac{(n-2)^2(n+1)}{2n-1} = \frac{\left(\frac{3^t - 3}{2}\right)^2 \left(\frac{3^t + 3}{2}\right)}{3^t}$
$= \frac{3^3 (3^{t-1} - 1)^2 (3^{t-1} + 1)}{8 \cdot 3^t}$.
Therefore, $t \leq 3$.
When $t=3$, $2n-1=27$, we get $n=14$.
At this point, $\frac{(n-2)^2(n+1)}{2n-1} = \frac{12^2 \cdot 15}{27}$ is an integer.