Let's suppose I know $\lambda_i$ to be an eigenvalue of the real negative definite square matrix $A$ of size $n$. $\lambda_i$ can either be real or complex (and $\lambda_{i+1}$ will then be its conjugate).
Is there a theorem or proof that could answer the question
For which $B$ is $\lambda_i$ an eigenvalue of $A+B$?
i.e.
Can a real matrix $B$ be found such that $det(A+B-\lambda_i I_n)=0 \;\;\; \forall i \in [1,n]$ ?
If your $n \times n$ matrix $A$ has $n$ distinct eigenvalues $\lambda_i$, what you're asking is that $A + B$ has the same characteristic polynomial as $A$. One possibility is $B = f(A) - A$ where $f$ is a polynomial that permutes the eigenvalues, i.e. $f(\lambda_i) = \lambda_{\sigma(i)}$, where $\sigma$ is a permutation of $\{1,\ldots,n\}$. Such an $f$ can be found by Lagrange interpolation.
By the way, if you're talking about a symmetric negative definite matrix (conventions differ on whether this is part of the definition), the eigenvalues are all real.