Question 1209918
Let $a_k = kx_k$ and $b_k = 1/k$.
Then the given condition is
$$\sum_{k=1}^n a_k = \sum_{k=1}^n kx_k = 1$$
We want to find the minimum value of
$$S = \sum_{k=1}^n \frac{x_k^2}{k} = \sum_{k=1}^n \frac{a_k^2}{k^3}$$

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

The minimum value of $\sum_{k=1}^n \frac{x_k^2}{k}$ is $\frac{1}{\sum_{k=1}^n k^3}$.
We know that $\sum_{k=1}^n k^3 = \left( \frac{n(n+1)}{2} \right)^2$.
So the minimum value is
$$\frac{1}{\left( \frac{n(n+1)}{2} \right)^2} = \frac{4}{n^2(n+1)^2}$$

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

Therefore, the minimum value of $\sum_{k=1}^n \frac{x_k^2}{k}$ is $\frac{1}{\sum_{k=1}^n k^3} = \frac{4}{n^2(n+1)^2}$.

Final Answer: The final answer is $\boxed{\frac{4}{n^2(n+1)^2}}$