jmc

Note that $$\begin{array}{rl}\frac{1}{1+{\omega }^k}+\frac{1}{1+{\omega }^{1997-k}}& =\frac{1}{1+{\omega }^k}+\frac{{\omega }^k}{{\omega }^k+{\omega }^{1997}}\\ & =\frac{1}{1+{\omega }^k}+\frac{{\omega }^k}{{\omega }^k+1}\\ & =\frac{1+{\omega }^k}{1+{\omega }^k}=1.\end{array}$$Thus, we can pair the terms $$\frac{1}{1+\omega },\frac{1}{1+{\omega }^2},\dots ,\frac{1}{1+{\omega }^{1995}},\frac{1}{1+{\omega }^{1996}}$$into $1996/2=998$ pairs, so that the sum of the numbers in each pair is 1. Also, $\frac{1}{1+{\omega }^{1997}}=\frac{1}{2},$ so the sum works out to $998+\frac{1}{2}=\boxed{\frac{1997}{2}}.$