II. (50 points) Given positive integers $x_{1}<x_{2}<\cdots<x_{n}$, and
$$
\begin{array}{l}
x_{1}+x_{2}+\cdots+x_{n}=2003, n \geqslant 2, \\
f(n)=n\left(x_{1}+x_{n}\right) .
\end{array}
$$

Find the minimum value of $f(n)$.
If $x_{1}>1$, let $x_{1}^{\prime}=1, x_{n}^{\prime}=x_{n}+x_{1}-1$, and the other $x_{i}$ values remain unchanged, then
$$
\begin{array}{l}
x_{1}^{\prime}+x_{2}+x_{3}+\cdots+x_{n-1}+x_{n}^{\prime}=2003, \\
f(n)=n\left(x_{1}+x_{n}\right)=n\left(1+x_{n}^{\prime}\right) .
\end{array}
$$

Assume $x_{1}=1$.
When $n=2$, $f(2)=4006$.
When $n \geqslant 3$, we have
$$
\begin{array}{l}
x_{2}<x_{3}<\cdots<x_{n} \\
\Rightarrow x_{n-1} \leqslant x_{n}-1, \cdots, x_{2} \leqslant x_{n}-(n-2) \\
\Rightarrow(n-1) x_{n}-[1+2+\cdots+(n-2)] \\
\quad \geqslant x_{2}+x_{3}+\cdots+x_{n}=2002 \\
\Rightarrow(n-1) x_{n}-\frac{1}{2}(n-1)(n-2) \geqslant 2002 \\
\Rightarrow x_{n} \geqslant \frac{1}{2}(n-2)+\frac{2002}{n-1} \\
\Rightarrow f(n)=n\left(1+x_{n}\right)=n+n x_{n} \\
\quad \geqslant n+\frac{1}{2} n(n-2)+\frac{2002 n}{n-1}=\frac{1}{2} n^{2}+\frac{2002 n}{n-1} .
\end{array}
$$

Let $g(n)=\frac{1}{2} n^{2}+\frac{2002 n}{n-1}$, and examine the monotonicity of $g(n)$:
When $n \leqslant 13$, $g(n)$ is monotonically decreasing;
When $n \geqslant 14$, $g(n)$ is monotonically increasing.
Also, $g(13)=2253+\frac{1}{3}, g(14)=2254$, therefore,
$$
f(n) \geqslant g(n) \geqslant 2253+\frac{1}{3} \Rightarrow f(n) \geqslant 2254 .
$$

When $n=14$, take
$$
x_{1}=1, x_{2}=148, x_{3}=149, \cdots, x_{14}=160 \text {, }
$$

then $f(n)_{\text {min }}=n\left(x_{1}+x_{n}\right)=2254$.