Let $n$ be a natural number. We can view the space of invertible symmetric matrices over a field as an open subset of$\mathbb A^{(n^2+n)/2}$. Inside the fourth power of this space, we have the closed subscheme consisting of tuples satisfying $A_1 + A_2 = A_3 + A_4$ and $A_1^{-1} + A_2^{-1} = A_3^{-1}+ A_4^{-1}$.

Is this subscheme a complete intersection of dimension $n^2+n$?

How many irreducible components does this subscheme have?

The motivation is that this would evaluate the fourth moment of symplectic Kloosterman sums, in the same way that Kloosterman's classical argument evaluates the fourth moment of the usual Kloosterman sums. However, I don't expect techniques from number theory to be helpful here.

The $n=1$ case has three irreducible components of dimension $2$, given by equations as follows $(x_3=x_1,x_4=x_2),(x_3=x_2,x_4=x_1),(x_2=-x_1,x_4=-x_3)$. Using these, we can make $3^n$ $2n$-dimensional families of diagonal examples. All of these are contained in an irreducible component of dimension at least $n^2+n$, as the scheme is defined by only $n^2+n$ equations in $2n^2 +2n$-space. However, only for a few obvious ones can I locate an $n^2+n$-dimensional family containing them.

Let me express what I think remains to be done after the two answers already given. Using Alex Gavrilov's algebraic formulation, I think we can classify all the irreducible components that arise when $A_1+A_2$ and $A_1^{-1}+A_2^{-1}$ are invertible. One might conjecture that all irreducible components arise on invertible irreducible component on one subspace and an irreducible component where $A_1+A_2=0$ on another subspace, as in David Speyer's answer. So the main thing to do that I don't know how to do is verify or disprove this conjecture.