5. (Problem 1500 from "Mathematics Bulletin") Given positive integers $x_{1}<x_{2}<\cdots<x_{n}, x_{1}+x_{2}+\cdots+$ $x_{n}=2003, n \geqslant 2$. Find the minimum value of $f(n)=n\left(x_{1}+x_{n}\right)$.
5. If $x_{1}>1$, let $x_{1}{ }^{\prime}=1, x^{\prime}{ }_{n}=x_{n}+x_{1}-1$, and the other $x_{i}$ values remain unchanged, then we have
$$
\begin{array}{l}
x^{\prime}{ }_{1}+x_{2}+\cdots+x_{n-1}+x^{\prime}{ }_{n}=2003 \\
f(n)=n\left(x_{1}+x_{n}\right)=n\left(1+x^{\prime}{ }_{n}\right)
\end{array}
$$

Therefore, we can assume $x_{1}=1$.
When $n=2$, $f(n)=2 \times 2003=4006$.
When $n \geqslant 3$, since $x_{2}4004$.
So when $n \leqslant 12$, $g(n)$ is monotonically decreasing, and when $n=13$, $g(n)$ is monotonically increasing.
Also, $g(12)=2256, g(13)=2253+\frac{1}{3}$,
Thus, $f(n) \geqslant g(n) \geqslant 2253+\frac{1}{3}$.
Therefore, $f(n) \geqslant 2254$.
Furthermore, when $n=12$, taking $x_{1}=1, x_{2}=177, x_{3}=178, \cdots, x_{12}=187, n\left(x_{1}+x_{2}\right)=2256$.
When $n=13$, taking $x_{1}=1, x_{2}=161, x_{3}=162, \cdots, x_{12}=171, x_{13}=174, n\left(x_{1}+x_{n}\right)=2275$.
When $n=14$, taking $x_{1}=1, x_{2}=148, x_{3}=149, \cdots, x_{14}=160, n\left(x_{1}+x_{n}\right)=14 \times 161=2254$.
Therefore, the minimum value of $f(n)=n\left(x_{1}+x_{n}\right)$ is 2254, at which $n=14, x_{1}=1, x_{2}=148, x_{3}=149, \cdots, x_{11}=160$.