jmc

To work with the inverse $f^{-1},$ we consider the range of each component of $f(x).$ Let $g(x)=x-2$ for $x<4,$ and let $h(x)=\sqrt{x}$ for $x\ge 4.$ For $x<4,$ we have $x-2<2,$ so the range of $g$ is the interval $(-\infty ,-2).$ For $x\ge 4,$ we have $\sqrt{x}\ge 2,$ so the range of $h$ is $[2,\infty ).$

The inverse of $g$ is $g^{-1}(x)=x+2,$ while the inverse of $h$ is $h^{-1}(x)=x^2.$ To compute $f^{-1}(x),$ we must use $g^{-1}$ if $x<2,$ and use $h^{-1}$ if $x\ge 2$: $$\begin{array}{rl}{f}^{-1}(-5)+f^{-1}(-4)+\cdots +f^{-1}(4)+f^{-1}(5)& =\left(g^{-1}(-5)+\cdots +g^{-1}(1)\right)+\left(h^{-1}(2)+\cdots +h^{-1}(5)\right)\\ & =\left((-3)+(-2)+\cdots +3\right)+\left(4+9+16+25\right)\\ & =0+54\\ & =\boxed{54}.\end{array}$$