I am wondering if there is a closed form expression for the sum \begin{align} \sum_{n=1}^{N-1} \frac{n}{N-n}. \end{align}
We note that there is a symmetry \begin{align} \sum_{n=1}^{N-1} \frac{n}{N-n} = \sum_{n=1}^{N-1}\frac{N-n}{n} \end{align} which might be helpful. One observation which I do not know may be helpful is that this can be seen as a special case of a general sum with Mobius conformal mappings \begin{align} z\rightarrow \gamma z=\frac{az+b}{cz+d}, \quad \gamma \in SL(2,\mathbb{C}), \quad \sum_{z \in \{1,\cdots,N-1\}} \gamma z \end{align} where $\gamma=\begin{pmatrix}N & -1 \\ 1 & 0\end{pmatrix}$. However, since the Mobius map is not linear, this may not be good way to start.
Please let me know if anyone knows the answer.