, 2019

The required minimum is $0$ and is achieved if and only if the $x_i$ are all equal to $1$. Let $x_1, \dots, x_n$ be positive real numbers satisfying the condition in the statement. Let $y_i = x_i/(x_i + n - 1)$, $i = 1, 2, \dots, n$, and notice that the $y_i$ are positive real numbers that add up to $1$. Express the $x_i$ in terms of the $y_i$ to get $x_i = (n-1)y_i/(1-y_i)$, and write successively
$$\begin{aligned} \sum_{i=1}^{n} \frac{1}{x_i} &= \frac{1}{n-1} \sum_{i=1}^{n} \frac{1-y_i}{y_i} = \frac{1}{n-1} \sum_{i=1}^{n} \frac{1}{y_i} \sum_{j \neq i} y_j = \frac{1}{n-1} \sum_{i \neq j} \frac{y_j}{y_i} = \frac{1}{n-1} \sum_{i \neq j} \frac{y_i}{y_j} \\ &= \frac{1}{n-1} \sum_{i=1}^{n} y_i \sum_{j \neq i} \frac{1}{y_j} \ge \frac{1}{n-1} \sum_{i=1}^{n} y_i \cdot \frac{(n-1)^2}{\sum_{j \neq i} y_j} = \sum_{i=1}^{n} \frac{(n-1)y_i}{1-y_i} = \sum_{i=1}^{n} x_i. \end{aligned}$$
Equality clearly forces the $y_i$ all equal to $1/n$, which is the case if and only if the $x_i$ are all equal to $1$.