Let $B$ and $-B$ be square complex matrices such that they are similar. If there is $m$ Jordan block $J_k(\lambda)$ in $B$, the Jordan block $J_k(-\lambda)$ also appears $m$ times in $B$.
This is my approach : First, since $B$ and $-B$ are similar, $B$ and $-B$ have the same Jordan canonical form. Also, if $\lambda$ is an eigenvalue of $B$, $-\lambda$ is also its eigenvalue. I claim that if $\lambda$ has multiplicity $m$, then $-\lambda$ also has multiplicity $m$ in $B$.
Let $\det(B - xI) = p(x), \det(-B - xI) = q(x)$. Then $$q(x) = (-1)^n \det(B - (-x)I) = (-1)^np(-x).$$
Assume that $p(x) = (x-\lambda)^a(x+\lambda)^bP(x)$ with $x-\lambda, x+\lambda \nmid P(x).$ Then $q(x) = (-1)^{n+a+b}(x+\lambda)^a(x-\lambda)^bP(-x).$ This should yield that $a=b$, that is, $\lambda, -\lambda$ has the same multiplicity in $B$.
How should I proceed form here to deduce that the multiplicity of $J_k(\lambda)$ and $J_k(-\lambda)$ is also the same in $B.$
If $N$ is nilpotent, then $N$ is similar to $-N$. If $\lambda I_k+J_k$ is a Jordan block with $\lambda\not= 0$, then it is similar to $\lambda I_k-J_k$.
Thus your matrix $B$ is similar to $diag(N,U_1,\cdots,U_r)$ where $N$ is nilpotent and the $U_i$ are in the form $diag(\lambda I_k+J_k,-\lambda I_k+J_k)$ where $\lambda \in spectrum(B)\setminus\{0\}$.
The essential argument is, for every $\lambda\in spectrum(B),l\in\mathbb{N}$, $dim(\ker(B-\lambda I_n)^l)=dim(\ker(-B-\lambda I_n)^l)$.