Solution: $1-\ln 2$
Rearranging the given equation gives $f^{\prime}(x) e^{-f(x)}=g^{\prime}(x) e^{-g(x)}$ for all $x$, so $\frac{d}{d x}\left(e^{-f(x)}-\right.$ $\left.e^{-g(x)}\right)=-f^{\prime}(x) e^{-f(x)}+g^{\prime}(x) e^{-g(x)}=0$. Thus, $e^{-f(x)}-e^{-g(x)}$ is a constant, and it must be less than $e^{-f(0)}=e^{-1}$. Thus, $e^{-f(2003)}$ $1-\ln 2$. On the other hand, we can find positive-valued functions $e^{-f(x)}, e^{-g(x)}$ that take on the required values at 0 and 2003 and have constant difference arbitrarily close to $e^{-1}$. For example, for arbitrarily large $t$, we can set $e^{-f(x)}=e^{-(t(2003-x)+1)}+e^{-1}-$ $e^{-(2003 t+1)}$ and $e^{-g(x)}=e^{-(t(2003-x)+1)}$, and we can check that the resulting functions $f, g$ satisfy the required conditions. Thus, we can make $f(2003)$ arbitrarily close to $1-\ln 2$, so this is the answer.
9. Two differentiable real functions $f(x)$ and $g(x)$ satisfy
$$
\frac{f^{\prime}(x)}{g^{\prime}(x)}=e^{f(x)-g(x)}
$$
for all $x$, and $f(0)=g(2003)=1$. Find the largest constant $c$ such that $f(2003)>c$ for all such functions $f, g$.