The minimum exists, as the summands are all non-negative integers; it is equal to $(n-1)^2$ and is achieved if, for instance, $x_1 = n$ and $x_2 = \cdots = x_n = n+1$; the verification is routine.

Let $s = x_1 + x_2 + \cdots + x_n$ and let $S$ denote the sum in the statement. Note that
$$\left\lfloor \frac{x_1 + \cdots + x_{k-1} + x_{k+1} + \cdots + x_n}{x_k} \right\rfloor = \left\lfloor \frac{s - x_k}{x_k} \right\rfloor = \left\lfloor \frac{s}{x_k} - 1 \right\rfloor = \left\lfloor \frac{s}{x_k} \right\rfloor - 1 \\ > \left( \frac{s}{x_k} - 1 \right) - 1 = \frac{s}{x_k} - 2, \quad k = 1, \dots, n.$$
Sum over $k = 1, 2, \dots, n$ to get
$$\begin{aligned} S &> s \cdot \left( \frac{1}{x_1} + \frac{1}{x_2} + \cdots + \frac{1}{x_n} \right) - 2n = \\ &= (x_1 + x_2 + \cdots + x_n) \left( \frac{1}{x_1} + \frac{1}{x_2} + \cdots + \frac{1}{x_n} \right) - 2n \ge n^2 - 2n, \end{aligned}$$
by the AM-HM (or Cauchy-Schwarz or Chebyshev) inequality.
Finally, as $S$ and $n^2 - 2n$ are both integers, $S \ge n^2 - 2n + 1 = (n-1)^2$, as desired. This ends the proof.