Let $F(x)=f(x)+x$, then $F(0)=1$.
Let $u$ be the smallest root of $f(x)=0$, then $F(u)=u$.
If $u0$
For any real number $x$, by (5) we have
$$\begin{aligned} 0 & =f(x) f(u)=f(x f(u)+u f(x)+x u) \\ & =f(u f(x)+u u) . \end{aligned}$$
$\therefore u f(x)+x u$ is a root of $f(x)=0$.
$\because u$ is the smallest root of $f(x)=0$,
$$\begin{array}{l} \therefore u f(x)+x u \geqslant u . \\ \text { Thus, } f(x)+x \geqslant 1, \end{array}$$

i.e., $f(x) \geqslant 1-x$.
$$\begin{array}{l} \therefore f(-1999) \geqslant 2000 . \\ \text { Also, } \because f(-1999) \leqslant 2000, \\ \therefore f(-1999)=2000 . \end{array}$$