One, (40 points) Given positive integers $n, k (n>k)$, and $x_{1}, x_{2}, \cdots, x_{k}>0$. Find $x_{k+1}, x_{k+2}, \cdots, x_{n}>0$, such that $\sum_{1 \leqslant i, j \leqslant n} \frac{x_{i}}{x_{j}}$ is minimized.
Let $A=\sum_{i=1}^{k} x_{i}, B=\sum_{i=1}^{k} \frac{1}{x_{i}}$.
$$
\begin{array}{l}
\text { Then } \sum_{1 \leqslant i, j \leqslant n} \frac{x_{i}}{x_{j}}=\left(\sum_{i=1}^{n} x_{i}\right)\left(\sum_{i=1}^{n} \frac{1}{x_{i}}\right)-n \\
=\left(A+\sum_{i=k+1}^{n} x_{i}\right)\left(B+\sum_{i=k+1}^{n} \frac{1}{x_{i}}\right)-n \\
\geqslant(\sqrt{A B}+n-k)^{2}-n .
\end{array}
$$

Equality holds if and only if $x_{k+1}=x_{k+2}=\cdots=x_{n}=\sqrt{\frac{A}{B}}$.