Two, the maximum value of $\sum_{i=1}^{2000} F_{i}$ is $\frac{2000}{3999}$.
Actually, since $x_{i} \in[0,1]$, first
$$\sum_{i=1}^{2000} F_{i} \leqslant \frac{\sum_{i=1}^{2000} x_{i}^{2000}}{\sum_{i=1}^{2000} x_{i}^{3999}+1999},$$

Let the right side of the inequality be $S$, we only need to prove $S \leqslant \frac{2000}{3999}$.
That is, $\sum_{i=1}^{2000} x_{i}^{2000} \leqslant \frac{2000}{3999}\left(\sum_{i=1}^{2000} x_{i}^{3999}+1999\right)$.
Let $f\left(x_{i}\right)=\frac{2000}{3999} x_{i}^{3999}-x_{i}^{2000}+\frac{1999}{3999}$.
(We only need to prove that for $\forall i=1,2, \cdots, 2000, f\left(x_{i}\right) \geqslant 0$)

And
$$\begin{array}{l} \text { 3999 } f(x)=2000 x^{3999}-3999 x^{2000}+1999 \\ =2000 x^{2000}\left(x^{1999}-1\right)-1999\left(x^{2000}-1\right) \\ =(x-1)\left[2000 x^{2000}\left(x^{1998}+x^{1997}+\cdots+1\right)\right. \\ \left.\quad-1999\left(x^{1999}+x^{1998}+\cdots+1\right)\right] \\ =(x-1)\left[\sum_{i=2000}^{3998}\left(x^{i}-1\right)\right. \end{array}$$
$$\begin{aligned} & +1999 \sum_{i=1}^{1999}\left(x^{i+1999}-x^{i}\right] \\ = & (x-1)^{2}\left[\sum_{i=2000}^{3998}\left(x^{i-1}+x^{i-2}+\cdots+1\right)\right. \\ & \left.+1999 \sum_{i=0}^{19988} x^{i} \sum_{i=1}^{1999} x^{i}\right] \end{aligned}$$
$\geqslant 0$.
Therefore, equation (1) holds. The condition for equality is
$$x_{1}=x_{2}=\cdots=x_{2000}=1 .$$