Let $ r $ be a root of the equation, so \[r^{98} + r^{97} + \dots + r + 1 = 0.\]Then \[(r - 1)(r^{98} + r^{97} + \dots + r + 1) = 0,\]which expands as $ r^{99} - 1 = 0 $. Hence, $ r^{99} = 1 $. Taking the absolute value of both sides, we get $ |r^{99}| = 1, $ so $ |r|^{99} = 1 $. Therefore, $ |r| = 1 $. We have shown that all roots lie on the unit circle. Hence, $ r \overline{r} = |r|^2 = 1 $ for any root $ r $. Since the polynomial $ x^{98} + x^{97} + x^{96} + \dots + x^2 + x + 1 $ has real coefficients, its nonreal roots come in conjugate pairs. Furthermore, if $ r $ is a root, then $ |r| = 1 $. If $ r $ is real, then the only possible values of $ r $ are 1 and $ -1, $ and neither of these are roots, so all the roots are nonreal, which means we can arrange all the roots in conjugate pairs. Without loss of generality, we can assume that $ \overline{r}_i = r_{99 - i} $ for $ 1 \le r \le 98 $. This also tells us that $ r_i r_{99 - i} = 1 $. Let \[S = \sum_{i = 1}^{98} \frac{r_i^2}{r_i + 1}.\]Then \begin{align*} 2S &= \sum_{i = 1}^{98} \left( \frac{r_i^2}{r_i + 1} + \frac{r_{99 - i}^2}{r_{99 - i} + 1} \right) \\ &= \sum_{i = 1}^{98} \left( \frac{r_i^2}{r_i + 1} + \frac{\frac{1}{r_i^2}}{\frac{1}{r_i} + 1} \right) \\ &= \sum_{i = 1}^{98} \left( \frac{r_i^2}{r_i + 1} + \frac{1}{r_i (r_i + 1)} \right) \\ &= \sum_{i = 1}^{98} \frac{r_i^3 + 1}{r_i (r_i + 1)} \\ &= \sum_{i = 1}^{98} \frac{r_i^2 - r_i + 1}{r_i} \\ &= \sum_{i = 1}^{98} \left( r_i - 1 + \frac{1}{r_i} \right).\end{align*}By Vieta's formulas, \[r_1 + r_2 + \dots + r_{98} = -1.\]Taking the conjugate, we get \[\overline{r}_1 + \overline{r}_2 + \dots + \overline{r}_{98} = -1,\]so \[\frac{1}{r_1} + \frac{1}{r_2} + \dots + \frac{1}{r_{98}} = -1.\]Therefore, $ 2S = -1 - 98 - 1 = -100, $ so $ S = \boxed{-50} $.