20. (Mathematics Competition Training Problems from "Middle School Mathematics Reference", Issue 10, 2004) Let the function $f(x)=\frac{a^{x}}{1+a^{x}}(a>$ $0, a \neq 1), [m]$ denote the greatest integer not exceeding the real number $m$ (i.e., the Gauss function), then the range of the function $\left[f(x)-\frac{1}{2}\right]+$ $\left[f(-x)-\frac{1}{2}\right]$ is $\qquad$ .
20. Fill $\{-1,0\}$. Reason: Let $g(x)=f(x)-\frac{1}{2}$, then $g(-x)=\frac{a^{-x}}{1+a^{-x}}-\frac{1}{2}=\frac{1}{a^{x}+1}-\frac{1}{2}=$ $-\left(\frac{a^{x}}{1+a^{x}}-\frac{1}{2}\right)=-g(x)$. Therefore, $\left[f(x)-\frac{1}{2}\right]+\left[f(-x)-\frac{1}{2}\right]=[g(x)]+[-g(x)]$.
Since $a^{x}>0$, then $0<f(x)=\frac{a^{x}}{1+a^{x}}<1$, thus $-\frac{1}{2}<g(x)<\frac{1}{2}$.
When $-\frac{1}{2}<g(x)<0$, $[g(x)]+[-g(x)]=-1+0=-1$;
When $g(x)=0$, $[g(x)]+[-g(x)]=0+0=0$;
When $0<g(x)<\frac{1}{2}$, $[g(x)]+[-g(x)]=0+(-1)=-1$.
In summary, the range of the function $\left[f(x)-\frac{1}{2}\right]+\left[f(-x)-\frac{1}{2}\right]$ is $\{-1,0\}$.