Solution. The following lemma can be proved by direct simplification.
Lemma. Let $S_{k}=a_{1}+a_{2}+\cdots+a_{k}$. Then
$$\sum_{k=1}^{n} a_{k} b_{k}=S_{n} b_{n}+\sum_{k=1}^{n-1} S_{k}\left(b_{k}-b_{k+1}\right) .$$

Let $S_{i}=x_{1}+x_{2}+\cdots+x_{i}$. By the given condition, $S_{n}=0$ and $\left|S_{i}\right| \leq \frac{1}{2}$ for $i=1, \cdots, n-1$. To see this, suppose $\left|S_{i}\right|>\frac{1}{2}$. Then $1=\left|x_{1}\right|+\left|x_{2}\right|+\cdots+$ $\left|x_{n}\right| \geq\left|x_{1}+\cdots+x_{i}\right|+\left|x_{i+1}+\cdots+x_{n}\right|=\left|S_{i}\right|+\left|-S_{i}\right|=2\left|S_{i}\right|>1$, which is a contradiction. By the lemma, we have
$$\sum_{k=1}^{n} \frac{x_{k}}{k}=S_{n} \cdot \frac{1}{n}+\sum_{k=1}^{n-1} S_{k}\left(\frac{1}{k}-\frac{1}{k+1}\right) .$$

Thus
$$\begin{aligned} \left|\sum_{k=1}^{n} \frac{x_{k}}{k}\right| & =\left|\sum_{k=1}^{n-1} S_{k}\left(\frac{1}{k}-\frac{1}{k+1}\right)\right| \leq \sum_{k=1}^{n-1}\left|S_{k}\right|\left(\frac{1}{k}-\frac{1}{k+1}\right) \\ & \leq \sum_{k=1}^{n-1} \frac{1}{2}\left(\frac{1}{k}-\frac{1}{k+1}\right)=\frac{1}{2}\left(1-\frac{1}{n}\right) . \end{aligned}$$

Second Solution. The inequality is achievable when $x_{1}= \pm \frac{1}{2}$ and $x_{n}=\mp \frac{1}{2}$ and the rest of $x_{i}=0$. So the inequality can be proved by the smoothing principle.
Let $a_{1} \geq \cdots \geq a_{k} \geq 0$ be the nonnegative terms among the $x_{i}$ 's and $b_{1} \leq$ $b_{2} \leq \cdots \leq b_{l}<0$ be the negative terms among the $x_{i}$ 's. Then we have $a_{1}+\cdots+a_{k}=1 / 2$ and $b_{1}+\cdots+b_{l}=-1 / 2$.
Without loss of generality, we can assume that the contribution from the nonnegative terms are greater than the contributions from the negative terms in the LHS. Note that for $0<i<j$, and $x, y \geq 0$, we have $x / i+y / j \leq$ $(x+y) / i+0 / j$. Applying this, we see that the LHS is less than or equal to
$$\frac{\sum_{i=1}^{k} a_{i}}{1}+\frac{0}{2}+\cdots+\frac{0}{n-1}+\frac{\sum_{i=1}^{k} b_{i}}{n}=\frac{1}{2}\left(1-\frac{1}{n}\right) .$$