Let $a, b$ and $c$ be three distinct positive integers. Consider the following $2k\times 2k$ matrix: \begin{equation} A=\begin{bmatrix} a&b&\cdots&b&c&c&\cdots&c\\ b&a&\cdots&b&c&c&\cdots&c\\ \vdots&&&\vdots&\vdots&&&\vdots\\ b&\cdots&b&a&c&c&\cdots&c\\ c&c&\cdots&c&a&b&\cdots&b\\ c&c&\cdots&c&b&a&\cdots&b\\ \vdots&&&\vdots&\vdots&&&\vdots\\ c&c&\cdots&c&b&b&\cdots&a \end{bmatrix}, \end{equation} where every diagonal entry equals $a$, every entry in the top right and bottom left $k\times k$ block equals $c$, and the remaining entries are $b$.
It can be checked that two of the eigenvalues of $A$ are $a+(k-1)b+kc$ and $a+(k-1)b-kc$, with eigenvectors $[1,\cdots,1]$ and $[1,\cdots,1,-1,\cdots,-1]$, respectively. Since this is a full rank matrix, it will have $2k$ nonzero eigenvalues. What can be said about the remaining eigenvalues?
Let $u,v\in\mathbb R^k$ be any pair of vectors orthogonal to $(1,1,\ldots,1)^T$. Then $$ A\pmatrix{u\\ v}=(a-b)\pmatrix{u\\ v}. $$ Those $u$s span a $(k-1)$-dimensional subspace and similarly for those $v$s. Therefore $a-b$ is an eigenvalue of geometric multiplicity $2(k-1)$.