19. Let $f(x)$ be a function that satisfies the following conditions:
(1) If $x>y$ and $f(x)+x \geqslant w \geqslant f(y)+y$, then there exists a real number $z \in[y, x]$ such that $f(z)=w-z$;
(2) The equation $f(x)=0$ has at least one solution, and among these solutions, there is one that is not greater than all the others;
(3) $f(0)=1$;
(4) $f(-2007) \leqslant 2008$;
(5) $f(x) f(y)=f[x f(y)+y f(x)+x y]$.

Find: the value of $f(-2007)$.
19. Let $F(x)=f(x)+x$, then $F(0)=1$. Suppose $u$ is the smallest root of $f(x)=0$, then $f(u)=0, F(u)=u$. If $u0$. For any real number $x$, by (5) we have
$$
0=f(x) f(u)=f[x f(u)+u f(x)+x u]=f[u f(x)+x u] .
$$

So $u f(x)+x u$ is a root of the equation $f(x)=0$.
Since $u$ is the smallest root of $f(x)=0$, we have
$$
u f(x)+x u \geqslant u .
$$

Thus
$$
f(x)+x \geqslant 1 .
$$

That is
$$
f(x) \geqslant 1-x .
$$

Therefore, $f(-2007) \geqslant 2008$. Also, $f(-2007) \leqslant 2008$, so $f(-2007)=2008$.