Arguing by contradiction, suppose that every sum of $k$ numbers from $x_1, x_2, \dots, x_n$ is strictly less than $k$. Then the numbers
$$a_j = x_j + x_{j+1} + \dots + x_{j+k-1}, \quad j = 1, 2, \dots, n$$
are also less than $k$ (where the indices from the sums $a_j$ are considered to be taken modulo $n$).
Adding up yields $a_1 + a_2 + \dots + a_n < nk$ and, since
$$a_1 + a_2 + \dots + a_n = k(x_1 + x_2 + \dots + x_n),$$
we get
$$x_1 + x_2 + \dots + x_n < n.$$

But $n^2 \le \left(\sum_{k=1}^n x_k\right) \left(\sum_{k=1}^n \frac{1}{x_k}\right) < n \sum_{k=1}^n \frac{1}{x_k}$, so $\sum_{k=1}^n \frac{1}{x_k} > n$. It follows that
$$\sum_{k=1}^{n} x_k = \sum_{k=1}^{n} \frac{1}{x_k^2} \ge \frac{1}{n} \left( \sum_{k=1}^{n} \frac{1}{x_k} \right)^2 > \frac{1}{n} \cdot n^2 = n,$$
contradiction.

Alternative Solution:
Suppose that $x_1 \le x_2 \le \dots \le x_n$; it is enough to prove that
$$x_{n-k+1} + \dots + x_{n-1} + x_n \ge k, \text{ for each } k=1, 2, \dots, n.$$