Example 2 Let the function $f(x)$ satisfy the following conditions:
(i) 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 $f(z)=$ $w-z$;
(ii) The equation $f(x)=0$ has at least one solution, and among the solutions of this equation, there is one solution that is not greater than all other solutions;
(iii) $f(0)=1$;
(iv) $f(-2003) \leqslant 2004$;
(v) $f(x) \cdot f(y)=f[x \cdot f(y)+y \cdot f(x)+x y]$.
Find the value of $f(-2003)$.
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 $u < 0$, then $F(u) = u < 0$, which contradicts $F(0) = 1$.
For any real number $x$, by condition (v), we have
$$
0 = f(x) \cdot f(u) = f[x \cdot f(u) + u \cdot f(x) + xu] = f[u \cdot f(x) + xu].
$$

Thus, $u \cdot f(x) + x \cdot u$ is a root of the equation $f(x) = 0$.
Since $u$ is the smallest root of $f(x) = 0$, then $u \cdot f(x) + x \cdot u \geqslant u$,
hence, $f(x) + x \geqslant 1$, i.e., $f(x) \geqslant 1 - x$.
Therefore, $f(-2003) \geqslant 2004$.
Also, by condition (iv), we have $f(-2003) \leqslant 2004$.
Thus, $f(-2003) = 2004$.