$A$ is a similar matrix of $B$, all eigenvalues of $A$ and $B$ are positive, $D$ is a positive semi-definite diagonal matrix. How to prove that the sign of all eigenvalues of $AD$ is always the same as that of $BD$?
For simplicity, we assume that $D$ is a diagonal matrix with $0$ or $1$ diagonal elements. I've done 100000 experiments and it's always true i.e. $AD$ always has the same numbers of positive eigenvalues as $BD$.
This is false: Take $$A=\left[\begin{array}{cc}2&0\\ 0&-1\end{array}\right],D=\left[\begin{array}{cc}1&0\\ 0&0\end{array}\right],~{\rm and~}B=\left[\begin{array}{cc}-1&0\\ 0&2\end{array}\right]=P^{-1}AP,$$ where $$P=\left[\begin{array}{cc}0&1\\ 1&0\end{array}\right].$$ Then $$AD=\left[\begin{array}{cc}2&0\\ 0&0\end{array}\right],BD=\left[\begin{array}{cc}-1&0\\ 0&0\end{array}\right].$$
[EDIT] The result is still false if one requires that all the eigenvalues are positive. Take $$A=\left[\begin{array}{cc}1&0\\ 0&2\end{array}\right],D=\left[\begin{array}{cc}1&0\\ 0&4\end{array}\right],~{\rm and~}B=\left[\begin{array}{cc}6&4\\ -5&-3\end{array}\right]=P^{-1}AP,$$ where $$P=\left[\begin{array}{cc}-5&-5\\ -5&-4\end{array}\right].$$ Then $$AD=\left[\begin{array}{cc}1&0\\ 0&8\end{array}\right],BD=\left[\begin{array}{cc}6&16\\ -5&-12\end{array}\right],$$ where $BD$ has eigenvalues $\{-2,-4\}.$