I am using a numerical method that involves a matrix $$ Z = X + Y, $$ where $X$ and $Y$ are both symmetric and invertible. The numerical method then proceeds to invert $Z$ for computations.
In practice, it is possible to invert $Z$ and work with it, but there are rare cases where the method blows up. However, it is not clear to me why would $X+Y$ be invertible and under which conditions.
So, under which conditions would $Z$ be invertible? What should I test for?
$-I$ is not positive, so I assume you mean that $X$ and $Y$ are only symmetric and invertible. If both of them are positive definite, then their sum is symmetric positive definite too, hence invertible.
In the simple case where $X = I$, it is clear that $X + Y$ is invertible if and only if $-1$ is not an eigenvalue of $Y$. Indeed, $\det(I + Y) = (-1)^n\chi_Y(-1)$ where $\chi_Y$ is the characteristic polynomial of $Y$. However, it is not easier to check with a machine if $-1$ is eigenvalue of a given matrix than to check if an other given matrix is invertible. Therefore, in this particular case, you may simply compute $\det(Z)$ and see if it vanishes or not. In the general case, it is even worse, I don't think there is a better way than simply add $X$ and $Y$ and then inverse their sum (when it is possible) like it is any symmetric matrix.
Maybe, you have more information about $X$ and $Y$ and you can find a better method depending on that (do they commute for example ?), but in this general setting, I believe there is no better solution.