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}
