18. We prove a stronger general conclusion: If real numbers $a_{1}, a_{2}, \cdots, a_{n+1}$, and $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{n}, a_{n+1}=a_{1}$, then
$$\sum_{k=1}^{n} f\left(a_{k+1}\right) a_{k} \leqslant \sum_{k=1}^{n} f\left(a_{k}\right) a_{k+1}$$

When $n=2$, the proposition is obviously true (since it is an equality). Assume the proposition holds for $n=m$. Then for $n=m+1$, for $b_{1} \geqslant b_{2} \geqslant \cdots \geqslant b_{m+1}, b_{m+2}=b_{1}$, we have
$$\begin{array}{l} \sum_{k=1}^{m+1} f\left(b_{k+1}\right) b_{k}-\sum_{k=1}^{m+1} f\left(b_{k}\right) b_{k+1}= \\ \sum_{k=1}^{m-1} f\left(b_{k+1}\right) b_{k}-\sum_{k=1}^{m-1} f\left(b_{k}\right) b_{k+1}+f\left(b_{m+1}\right) b_{m}+f\left(b_{1}\right) b_{m+1}-f\left(b_{m}\right) b_{m+1}-f\left(b_{m+1}\right) b_{1}= \\ {\left[\sum_{k=1}^{m-1} f\left(b_{k+1}\right) b_{k}+f\left(b_{1}\right) b_{m}-\sum_{k=1}^{m-1} f\left(b_{k}\right) b_{k+1}-f\left(b_{m}\right) b_{1}\right]+} \\ {\left[\left(b_{1}-b_{m+1}\right) f\left(b_{m}\right)-\left(b_{1}-b_{m}\right) f\left(b_{m+1}\right)-\left(b_{m}-b_{m+1}\right)-f\left(b_{1}\right)\right]} \end{array}$$

By the assumption, the first polynomial is less than or equal to 0. By the convexity of $f(x)$, the second polynomial is also less than or equal to 0. Therefore, when $n=m+1$, the proposition holds. In summary, this conclusion is true. When $n=2003$, it is the problem at hand.