Consider a $N \times N$ matrix $A=B_1+B_2$, where $B_1$ is a diagonal matrix with all the diagonal entries between $0$ and $1$ and $B_2$ is a skew symmetric matrix which can be written as
$$B_2= \begin{bmatrix} 0_{N-1} & f \\ -f^T & 0 \end{bmatrix}$$
where $0_{N-1}$ is the $N-1 \times N-1$ zero matrix, $f$ is a $N-1$ by $1$ vector. So $B_2=-B_2^T$.
My question is how many complex eigenvalues (non-zero imaginary part) $A$ has?
I have run some numerical simulations. It seems that $A$ can at most have one pair of complex eigenvalues. And it is plausible to me because the skew symmetric $B_2$ is only rank $2$.
How can we prove this mathematically or come up with a counter example? If needed, we can put an upper bound on $|f|_2$. But I don't know that if it is necessary, as indicated from my simulations.
We only need $B_1$ to be diagonal. Positive definiteness is not needed. Let $A=\pmatrix{D&f\\ -f^T&a}$. If some entries of $f$ is zero, we can reduce the problem to case with a smaller-sized $A$. Assume $f$ is entrywise nonzero. As Friedrich Philipp suggests in his comment, we may make use of Schur complement: \begin{align} \det(tI-A)&=\det(tI-D)\left(t-a+f^T(tI-D)^{-1}f)\right)\\ &=\det(tI-D)\left(t-a+\sum_{j=1}^{n-1} \frac{f_j^2}{t-d_j}\right). \end{align} Hence it suffices to show that $g(t)=t-a+\sum_{j=1}^{n-1} \frac{f_j^2}{t-d_j}$ has at least $n-2$ real roots.
By permutation, we may assume that the diagonal entries are arranged in strictly ascending order. Since $g(d_j^-)=-\infty$ and $g(d_j^+)=+\infty$, if the diagonal elements of $D$ are distinct, $g$ changes sign on every interval $(d_j,d_{j+1})$. Consequently, $g$ changes sign at least $n-2$ times on $\mathbb R\setminus\{d_1,d_2,\ldots,d_{n-1}\}$ and $A$ has at least $n-2$ (distinct) real eigenvalues. By continuity of eigenvalues in the matrix entries, $A$ also has at least $n-2$ real eigenvalues when some diagonal elements of $D$ are the same.