SELECTION TESTS FOR THE 2019 BMO AND IMO

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 $$\begin{array}{rl}\sum _{i=1}^n\frac{1}{x_i}& =\frac{1}{n-1}\sum _{i=1}^n\frac{1-y_i}{y_i}=\frac{1}{n-1}\sum _{i=1}^n\frac{1}{y_i}\sum _{j\ne i}{y}_j=\frac{1}{n-1}\sum _{i\ne j}\frac{y_j}{y_i}=\frac{1}{n-1}\sum _{i\ne j}\frac{y_i}{y_j}\\ & =\frac{1}{n-1}\sum _{i=1}^n{y}_i\sum _{j\ne i}\frac{1}{y_j}\ge \frac{1}{n-1}\sum _{i=1}^n{y}_i\cdot \frac{(n-1{)}^2}{{\sum }_{j\ne i}{y}_j}=\sum _{i=1}^n\frac{(n-1)y_i}{1-y_i}=\sum _{i=1}^n{x}_i.\end{array}$$ 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 $1$.