Let $g\left(x\right)=1+x-\left[x\right]$ and $f\left(x\right)=\left\{\begin{array}{l}-1 if x<0\\ 0 if x=0\\ 1 if x>0\end{array}\right.$ then $\frac{f\left(g\left(2009\right)\right)}{g\left(f\left(2009\right)\right)}$

The given functions are $g\left(x\right)=1+x-\left[x\right]$ and $f\left(x\right)=\left\{\begin{array}{l}-1 if x<0\\ 0 if x=0\\ 1 if x>0\end{array}\right.$

Consider $f\left(g\left(2009\right)\right)=f\left(1\right)=1$

and $g\left(f\left(2009\right)\right)=g\left(1\right)=1$

Therefore, $\frac{f\left(g\left(2009\right)\right)}{g\left(f\left(2009\right)\right)}=\frac{1}{1}=\boxed{1}$