One. (40 points) Let $n(n>1)$ be a given positive integer, and $x_{1}, x_{2}, \cdots, x_{n}$ be any real numbers, satisfying $\sum_{i=1}^{n} x_{i}^{2}=1$. Find the maximum value of $\sum_{k=1}^{n} k x_{k}^{2}+\sum_{1 \leqslant i<j \leqslant n}(i+j) x_{i} x_{j}$.
$$
\begin{array}{l}
\sum_{k=1}^{n} k x_{k}^{2}+\sum_{1 \leqslant i<j \leqslant n}(i+j) x_{i} x_{j} \\
=\left(x_{1}+x_{2}+\cdots+x_{n}\right)\left(x_{1}+2 x_{2}+\cdots+n x_{n}\right) \\
=\frac{1}{\lambda}\left(\lambda x_{1}+\lambda x_{2}+\cdots+\lambda x_{n}\right)\left(x_{1}+2 x_{2}+\cdots+n x_{n}\right) \\
\leqslant \frac{1}{\lambda}\left(\frac{(\lambda+1) x_{1}+\cdots+(\lambda+n) x_{n}}{2}\right)^{2} \\
\leqslant \frac{1}{4 \lambda}\left(x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}\right) \cdot \\
\quad\left((1+\lambda)^{2}+(2+\lambda)^{2}+\cdots+(\lambda+n)^{2}\right) \\
=\frac{1}{4 \lambda}\left((1+\lambda)^{2}+(2+\lambda)^{2}+\cdots+(n+\lambda)^{2}\right) .
\end{array}
$$

The condition for equality in the above equation is
$$
\begin{array}{l}
\frac{x_{1}}{1+\lambda}=\frac{x_{2}}{2+\lambda}=\cdots=\frac{x_{n}}{n+\lambda}, \\
\lambda\left(x_{1}+x_{2}+\cdots+x_{n}\right)=x_{1}+2 x_{2}+\cdots+n x_{n} . \\
\text { Let } \lambda=\sqrt{\frac{1^{2}+2^{2}+\cdots+n^{2}}{n}} \\
=\sqrt{\frac{(n+1)(2 n+1)}{6}},
\end{array}
$$

that is, $x_{i}=(i+\lambda) \sqrt{\frac{1}{\sum_{i=1}^{n}(i+\lambda)^{2}}}$ satisfies the condition for equality. At this point, the maximum value sought is
$$
\frac{n}{4}\left(n+1+2 \sqrt{\frac{(n+1)(2 n+1)}{6}}\right) .
$$