jmc

In order to find $f^{-1}$ we substitute $f^{-1}(x)$ into our expression for $f$. This gives $$f(f^{-1}(x))=2f^{-1}(x)+1.$$Since $f(f^{-1}(x))=x$, this equation is equivalent to $$x=2f^{-1}(x)+1,$$which simplifies to $$f^{-1}(x)=\frac{x-1}{2}.$$If we assume $x$ solves $f^{-1}(x)=f(x^{-1})$, then we get $$\frac{x-1}{2}=\frac{2}{x}+1=\frac{2+x}{x}.$$Cross-multiplying gives $$x^2-x=4+2x.$$Then $x^2-3x-4=0$. Factoring gives $(x-4)(x+1)=0$, from which we find $x=4$ or $x=-1$. The sum of the solutions is $4+(-1)=\boxed{3}$.

Alternatively, since Vieta's formula tells us that the sum of the roots of a quadratic $a{x}^2+bx+c$ is $-\frac{b}{a}$, the sum of the roots of $x^2-3x-4$ is $-\frac{-3}{1}=\boxed{3}$.