I am looking for the values of $Z$ for which the determinant of the following $N$-dimensional matrix vanishes:
\begin{equation} \begin{bmatrix} N(1-Z) & N-1 & N-2 & \cdots & \cdots & \cdots & \cdots & 1 \\ N-1 &(N-1)(1-Z) & N-2 & \ddots & && & \vdots \\ N-2 & N-2 & (N-2)(1-Z) & \ddots & \ddots & & & \vdots \\ \vdots & \ddots & \ddots & \ddots & \ddots & \ddots & & \vdots \\ \vdots & & \ddots & \ddots & \ddots & \ddots & \ddots& \vdots\\ \vdots & & & \ddots & \ddots & 3(1-Z) & 2 & 1\\ \vdots & && & \ddots & 2 & 2(1-Z) & 1\\ 1 & \cdots & \cdots & \cdots & \cdots & 1 & 1 & 1-Z \\ \end{bmatrix} \end{equation}
From physical considerations I know that all solutions for $Z$ are real and I have also proved that the sum of all solutions equals $N$. Does there exist a simple way to compute the solutions? I have reasons to believe that the solution is simple, because this was a contest problem.
Consider $n\times n$ matrix $A$: \begin{align} a_{11}&=n;\\ a_{1,i}&=a_{1,i-1}-1;\quad a_{i,1}=a_{i-1,1}-1;\quad\forall i=2\dots n; \\ a_{ij}&=a_{i-1,j-1}-1 \quad\forall i,j=2\dots n \end{align}
and $n\times n$ diagonal matrix $M$: $m_{ij}=0,\quad i\ne j;\quad m_{ii}=a_{ii},\quad i=1\dots n.$
Then the original somewhat strange-looking problem \begin{align} |A-M z| =0 \end{align} can be transformed easily into familiar eigenvalue problem, either $|(M^{-1}A)-z I|=0$ or $|(A^{-1}M)-\zeta I|=0,\quad \zeta=\frac{1}{z}$, whichever would be simpler to solve.
It looks like matrix $C=A^{-1}M$ is a better candidate, since $B=A^{-1}$ has a trivial tridiagonal structure: $b_{11}=1$, $2$ for all the rest of diagonal elements, while all elements above and below main diagonal are $-1$.
The matrix $A^{-1}M$ has also a well-defined tridiagonal structure, thus $|(A^{-1}M)-\zeta I|=0$ is much simpler to solve than the original system (Edit):
\begin{align} C_n&=A_n^{-1}M_n= \left[\begin{matrix} n & -(n-1) & & & \\ -n & 2(n-1) & -(n-2) & & \\ & \cdots & \cdots\quad\cdots & & \\ & & \cdots & \cdots\quad\cdots & \\ & & -3 & 2\cdot2 & -1 \\ & & & -2 & 2\cdot1 \end{matrix}\right] \end{align}
\begin{align} |C_1-\zeta I_1|&=1-\zeta, \\ |C_2-\zeta I_2|&=2-4 \zeta+\zeta^2, \\ |C_3-\zeta I_3|&=6-18\zeta+9\zeta^2-\zeta^3, \\ |C_4-\zeta I_4|&=24-96\zeta+72\zeta^2-16\zeta^3+\zeta^4, \\ |C_5-\zeta I_5|&=120-600\zeta+600\zeta^2-200\zeta^3+25\zeta^4-\zeta^5. \end{align}
Thus the eigenvalues of the matrix $C_n$ are the roots $\zeta_1,\dots,\zeta_n$ of the Laguerre polynomial
\begin{align} n! L_n(\zeta)&= \sum_{m=0}^{n} a_m (-\zeta)^m, \\ a_m&={n\choose m}^2 (n-m)! \end{align}
Finally, original roots $z_i=1/\zeta_i$, and indeed, $\sum_{i=1}^n z_i=n$:
\begin{align} \sum_{i=1}^n z_i&= \frac{a_1}{a_0} = \frac{n^2\cdot (n-1)!}{n!}=n. \end{align}
Other interesting facts about $z_i$ are:
\begin{align} \sum_{i=1}^n \frac{1}{z_i}&= a_{n-1} = {n\choose n-1}^2=n^2;\qquad \prod_{i=1}^n z_i = \frac{1}{a_0} = \frac{1}{n!}. \end{align}