This is not really an answer, but hopefully may be of help. As David Speyer already pointed out, the problem is basically about a matrix equation $$(X+Y+Z)^{-1}=X^{-1}+Y^{-1}+Z^{-1}.$$ Three symmetric matrices make a space of dimension $\frac{3}{2}n(n+1)$, and the above equation makes a variety of codimension $\frac{1}{2}n(n+1)$ in it. So, I would agree that the irreducible component of the largest dimension has dimension $n^2+n$.

Note that if you multiply each matrix by (the same) scalar, then it will still be a solution of the same equation. So, it is natural to consider this as a projective variety (a subvariety in $\mathbb{P}^{\frac{3}{2}n(n+1)-1}$). (Of course, it is a closure, i.e. it has points which do not correspond to triples of invertible matrices.)

Back to the original question. Denote $A=Y+Z, B=Y^{-1}+Z^{-1}$. Then we have an equation for $X$, $$A+XBX+ABX=0.$$ (And, using conjugation, $ABX=XBA$.) Equations of this sort are known as algebraic Riccati equations (though this one is a bit special). If I am not mistaken, for a general $Y,Z$ all of its solutions are Galois conjugate, which means that there is only one irreducible component of maximal dimension. The case $n=1$ (when there are three components) must be an exception.

[EDIT] (The above is almost true except there are other components.) Consider first the case when $Y$ and $Z$ are general. We may chose a square root $B^{1/2}$ and consider matrices $$x=B^{1/2}XB^{1/2},\,\,a=B^{1/2}AB^{1/2}.$$ In this terms, the above equation is $$xa=ax,\,x^2+ax+a=0.$$ In the general case, it has $2^n$ solutions (consider $a$ in the diagonal form.) I presumed that all these solutions are Galois conjugate, but on second thought, no. At least, there are two obvious solutions $X=-Y$ and $X=-Z$ which are not conjugate to anything. For $n>1$ there are more solutions. Apparently, the simplest possible picture here is four irreducible components, $X+Y=0, X+Z=0, Y+Z=0$ and the fourth one comes from extra solutions of the above equation (for $n>1$). Not sure if it is a correct picture though.

This is not really an answer, but hopefully may be of help. As David Speyer already pointed out, the problem is basically about a matrix equation $$(X+Y+Z)^{-1}=X^{-1}+Y^{-1}+Z^{-1}.$$ Three symmetric matrices make a space of dimension $\frac{3}{2}n(n+1)$, and the above equation makes a variety of codimension $\frac{1}{2}n(n+1)$ in it. So, I would agree that the irreducible component of the largest dimension has dimension $n^2+n$.

Note that if you multiply each matrix by (the same) scalar, then it will still be a solution of the same equation. So, it is natural to consider this as a projective variety (a subvariety in $\mathbb{P}^{\frac{3}{2}n(n+1)-1}$). (Of course, it is a closure, i.e. it has points which do not correspond to triples of invertible matrices.)

Back to the original question. Denote $A=Y+Z, B=Y^{-1}+Z^{-1}$. Then we have an equation for $X$, $$A+XBX+ABX=0.$$ (And, using conjugation, $ABX=XBA$.) Equations of this sort are known as algebraic Riccati equations (though this one is a bit special). If I am not mistaken, for a general $Y,Z$ all of its solutions are Galois conjugate, which means that there is only one irreducible component of maximal dimension. The case $n=1$ (when there are three components) must be an exception.

[EDIT] (The above is almost true except there are other components.) Consider first the case when $Y$ and $Z$ are general. We may chose a square root $B^{1/2}$ and consider matrices $$x=B^{1/2}XB^{1/2},\,\,a=B^{1/2}AB^{1/2}.$$ In this terms, the above equation is $$xa=ax,\,x^2+ax+a=0.$$ In the general case, it has $2^n$ solutions (consider $a$ in the diagonal form.) I presumed that all these solutions are Galois conjugate, but on second thought, no. At least, there are two obvious solutions $X=-Y$ and $X=-Z$ which are not conjugate to anything. For $n>1$ there are more solutions.

So, there are four components, $X+Y=0, X+Z=0, Y+Z=0$ and the fourth one (probably reducible) comes from extra solutions of the above equation (for $n>1$). So, it is not much. Still, the above argument has a point: it should be possible to count the components once you figure out how to divide these $2^n$ solutions between them.

This is not really an answer, but hopefully may be of help. As David Speyer already pointed out, the problem is basically about a matrix equation $$(X+Y+Z)^{-1}=X^{-1}+Y^{-1}+Z^{-1}.$$ Three symmetric matrices make a space of dimension $\frac{3}{2}n(n+1)$, and the above equation makes a variety of codimension $\frac{1}{2}n(n+1)$ in it. So, I would agree that the irreducible component of the largest dimension has dimension $n^2+n$.

Note that if you multiply each matrix by (the same) scalar, then it will still be a solution of the same equation. So, it is natural to consider this as a projective variety (a subvariety in $\mathbb{P}^{\frac{3}{2}n(n+1)-1}$). (Of course, it is a closure, i.e. it has points which do not correspond to triples of invertible matrices.)

Back to the original question. Denote $A=Y+Z, B=Y^{-1}+Z^{-1}$. Then we have an equation for $X$, $$A+XBX+ABX=0.$$ (And, using conjugation, $ABX=XBA$.) Equations of this sort are known as algebraic Riccati equations (though this one is a bit special). If I am not mistaken, for a general $Y,Z$ all of its solutions are Galois conjugate, which means that there is only one irreducible component of maximal dimension. The case $n=1$ (when there are three components) must be an exception.

This is not really an answer, but hopefully may be of help. As David Speyer already pointed out, the problem is basically about a matrix equation $$(X+Y+Z)^{-1}=X^{-1}+Y^{-1}+Z^{-1}.$$ Three symmetric matrices make a space of dimension $\frac{3}{2}n(n+1)$, and the above equation makes a variety of codimension $\frac{1}{2}n(n+1)$ in it. So, I would agree that the irreducible component of the largest dimension has dimension $n^2+n$.

Note that if you multiply each matrix by (the same) scalar, then it will still be a solution of the same equation. So, it is natural to consider this as a projective variety (a subvariety in $\mathbb{P}^{\frac{3}{2}n(n+1)-1}$). (Of course, it is a closure, i.e. it has points which do not correspond to triples of invertible matrices.)

Back to the original question. Denote $A=Y+Z, B=Y^{-1}+Z^{-1}$. Then we have an equation for $X$, $$A+XBX+ABX=0.$$ (And, using conjugation, $ABX=XBA$.) Equations of this sort are known as algebraic Riccati equations (though this one is a bit special). If I am not mistaken, for a general $Y,Z$ all of its solutions are Galois conjugate, which means that there is only one irreducible component of maximal dimension. The case $n=1$ (when there are three components) must be an exception.

[EDIT] (The above is almost true except there are other components.) Consider first the case when $Y$ and $Z$ are general. We may chose a square root $B^{1/2}$ and consider matrices $$x=B^{1/2}XB^{1/2},\,\,a=B^{1/2}AB^{1/2}.$$ In this terms, the above equation is $$xa=ax,\,x^2+ax+a=0.$$ In the general case, it has $2^n$ solutions (consider $a$ in the diagonal form.) I presumed that all these solutions are Galois conjugate, but on second thought, no. At least, there are two obvious solutions $X=-Y$ and $X=-Z$ which are not conjugate to anything. For $n>1$ there are more solutions. Apparently, the simplest possible picture here is four irreducible components, $X+Y=0, X+Z=0, Y+Z=0$ and the fourth one comes from extra solutions of the above equation (for $n>1$). Not sure if it is a correct picture though.