Note that
\begin{align*}
\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{align*}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 $998 + \frac{1}{2} = \boxed{\frac{1997}{2}}.$
Question:
If $\omega^{1997} = 1$ and $\omega \neq 1,$ then evaluate
\[\frac{1}{1 + \omega} + \frac{1}{1 + \omega^2} + \dots + \frac{1}{1 + \omega^{1997}}.\]
Answer:
Let's think step by step.

user
\frac{1997}{2}
If $\omega^{1997} = 1$ and $\omega \neq 1,$ then evaluate
\[\frac{1}{1 + \omega} + \frac{1}{1 + \omega^2} + \dots + \frac{1}{1 + \omega^{1997}}.\]
test