$$\begin{aligned}
\sum_{i = 1}^{1999}{\frac{x_i^3}{x_i^2 + 2x_i + 4}} - 8\sum_{i = 1}^{1999}{\frac{1}{x_i^2 + 2x_i + 4}} &= \sum_{i = 1}^{1999}{\frac{x_i^3 - 8}{x_i^2 + 2x_i + 4}}\\\\
&= \sum_{i = 1}^{1999}{\frac{(x_i - 2)(x_i^2 + 2x_i + 4)}{x_i^2 + 2x_i + 4}}\\\\
&= \sum_{i = 1}^{1999}{(x_i - 2)}\\\\
&=  - 2 \times 1999 + \sum_{i = 1}^{1999}{x_{i}}
\end{aligned}$$
より，求める値は $\textbf{4718}$ と計算できる．
公式解説
https://onlinemathcontest.com/contests/omc105/editorial/3282
　$1999$ 個の実数 $x_1, x_2, \cdots, x_{1999}$ は次を満たします．
$$\sum_{i = 1}^{1999}{\frac{1}{x_i^2 + 2x_i + 4}} = 160,\quad \sum_{i = 1}^{1999}{\frac{x_i^3}{x_i^2 + 2x_i + 4}} = 2000$$
このとき, $x_1 + x_2 + \cdots + x_{1999}$ の値を求めて下さい．