The number $a_k$ of the multiples of $k$ in the set $\{1, 2, \dots, n\}$ is $\lfloor \frac{n}{k} \rfloor$. Let us count the appearances of the term $\frac{1}{i}$ in the sum $x_1 + x_2 + \dots + x_n$. The term $\frac{1}{i}$ belongs in the sum $x_k = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{a_k}$ if and only if $i \le a_k = \lfloor \frac{n}{k} \rfloor$.

Fix $i = 1, 2, \dots, n$ and denote by $m = m(i)$ the index of the last sum $x_m$ containing $\frac{1}{i}$. Then $i \le \lfloor \frac{n}{m} \rfloor \le \frac{n}{m}$, i.e. $m \le \frac{n}{i}$.

Remember $\frac{1}{i}$ belongs to the sums $x_1, x_2, \dots, x_m$ and those only, therefore one has
$$x_1 + x_2 + \dots + x_n = \sum_{i=1}^{n} \frac{1}{i} m(i) \le \sum_{i=1}^{n} \frac{1}{i} \cdot \frac{n}{i} = n \sum_{i=1}^{n} \frac{1}{i^2}.$$
The conclusion immediately ensues.