Applying AM-GM gives
$$\begin{aligned} \left(\sum_{k=1}^{n} \frac{x_k}{k}\right) \left(\sum_{k=1}^{n} kx_k\right) &= \frac{1}{n} \cdot \left(\sum_{k=1}^{n} \frac{nx_k}{k}\right) \left(\sum_{k=1}^{n} kx_k\right) \le \\ &\le \frac{1}{n} \cdot \frac{1}{4} \left(\sum_{k=1}^{n} \frac{nx_k}{k} + \sum_{k=1}^{n} kx_k\right)^2 = \\ &= \frac{1}{4n} \left(\sum_{k=1}^{n} x_k \left(\frac{n}{k} + k\right)\right)^2 \le \frac{(n+1)^2}{4n} \left(\sum_{k=1}^{n} x_k\right)^2. \end{aligned}$$
(The last inequality is proved by $\frac{n}{k} + k \le n + 1$, as it is equivalent to $(n-k)(k-1) \ge 0$.)
This gives us the necessary upper bound; this bound is achieved for instance if $x_1 = x_n = 1$ and $x_2 = \dots = x_{n-1} = 0$.

For the lower bound, estimate the numerator by Cauchy-Schwarz inequality:
$$\left(\sum_{k=1}^{n} \frac{x_k}{k}\right) \left(\sum_{k=1}^{n} kx_k\right) \ge \left(\sum_{k=1}^{n} \sqrt{\frac{x_k}{k}} \cdot \sqrt{kx_k}\right)^2 = \left(\sum_{k=1}^{n} x_k\right)^2;$$
the equality holds here if exactly one of $x_i$s is non-zero.