For $n \ge 8$ the number $2(n-2)+1$ never divides $(n-2)(n-3)/2$.
Any ideas how to prove this? I see that $(n-2)(n-3)/2 = 1 + 2 + \ldots + (n-3)$. If I suppose that $2(n-2)+1$ divides $(n-2)(n-3)/2$ then it should also divide their difference \begin{align*} (n-2)(n-3)/2 - (2(n-2) + 1) & = \sum_{k=1}^{n-3} k - (2n - 3) \\ & = \sum_{k=1}^{n-4} k + (n-3) - 2n + 3 \\ & = \sum_{k=1}^{n-4} k - n \\ & = \frac{(n-3)(n-4)}{2} - n. \end{align*} That is all I have, hope you can give me some hints!?
Note that $2n-3$ is odd, so it is enough to show it does not divide $4(n-2) (n-3)$. which is $4n^2-20n+24$. Do a polynomial division. We get $2n-7+\frac{3}{2n-3}$.
If $2n-3\gt 3$, then $\frac{3}{2n-3}$ is not an integer.
Another way: Let $n\gt 3$. Note that $2(n-2)+1$ and $n-2$ are relatively prime. So it is enough to show that $2n-3$ cannot divide $n-3$. This is obvious.
A Puzzle: We have given two straightforward arguments that we cannot have divisibility if $n\gt 3$. How did they come up with the weaker $\ge 8$?