4. Real numbers $a, b, c$ satisfy $a+b>0, b+c>0$, $c+a>0, f(x)$ is an odd function on $\mathbf{R}$, and is a strictly decreasing function, i.e., if $x_{1}<x_{2}$ then $f(x_{1})>f(x_{2})$
(D) $f(a)+2 f(b)+f(c)=2004$
4.B.
$x \in \mathbf{R}, f(x)$ is an odd function, so $f(0)=0$. Since $f(x)$ is a strictly decreasing function, for $x>0$, we have $f(x)<0$, then
$$
\begin{array}{l}
a>-b \Rightarrow f(a)<f(-b)=-f(b) \\
\Rightarrow f(a)+f(b)<0 .
\end{array}
$$

Similarly, $f(b)+f(c)<0, f(c)+f(a)<0$.
Adding them up, we get $2[f(a)+f(b)+f(c)]<0$.