Question 1209919
Let $a_k = x_k \sqrt{k}$ for $k = 1, 2, \dots, n$.
Then the given condition becomes
$$\sum_{k=1}^n x_k^2 k = \sum_{k=1}^n a_k^2 = 1$$

We want to find the maximum value of
$$S = \left( \sum_{k=1}^n \frac{x_k}{k} \right)^2 = \left( \sum_{k=1}^n \frac{a_k}{k\sqrt{k}} \right)^2$$

By the Cauchy-Schwarz inequality, we have
$$\left( \sum_{k=1}^n \frac{a_k}{k\sqrt{k}} \right)^2 \le \left( \sum_{k=1}^n a_k^2 \right) \left( \sum_{k=1}^n \frac{1}{k^3} \right)$$
Since $\sum_{k=1}^n a_k^2 = 1$, we have
$$S \le \sum_{k=1}^n \frac{1}{k^3}$$

The maximum value of $\left( \sum_{k=1}^n \frac{x_k}{k} \right)^2$ is $\sum_{k=1}^n \frac{1}{k^3}$.

To achieve equality in Cauchy-Schwarz inequality, we need
$$\frac{a_1}{1\sqrt{1}} = \frac{a_2}{2\sqrt{2}} = \dots = \frac{a_n}{n\sqrt{n}} = c$$
where $c$ is a constant.
Then $a_k = ck\sqrt{k}$, so $x_k \sqrt{k} = ck\sqrt{k}$, which means $x_k = ck$.
Substituting into the given condition, we have
$$\sum_{k=1}^n kx_k^2 = \sum_{k=1}^n k(ck)^2 = c^2 \sum_{k=1}^n k^3 = 1$$
$$c^2 = \frac{1}{\sum_{k=1}^n k^3}$$
$$c = \frac{1}{\sqrt{\sum_{k=1}^n k^3}}$$
Then $x_k = \frac{k}{\sqrt{\sum_{k=1}^n k^3}}$.

Therefore, the maximum value of $\left( \sum_{k=1}^n \frac{x_k}{k} \right)^2$ is $\sum_{k=1}^n \frac{1}{k^3}$.

Final Answer: The final answer is $\boxed{\sum_{k=1}^n \frac{1}{k^3}}$