$n \ge 2$. We show the minimum value is $n-1$: Rewrite the condition as $\sum\frac {x_i^2-n+1}{1+x_i^2}=0$. Note that if $x_i \ge x_j$, $\frac{x_i^2-n-1}{1+x_i^2}-\frac{x_j^2-n-1}{1+x_j^2}=\frac{(n+2)(x_i^2-x_j^2)}{(1+x_i^2)(1+x_j^2)}\ge 0$. For $x_i,x_j$, $\frac {1}{1+x_i^2}+\frac {1}{1+x_j^2} \leq 1 \implies x_i^2x_j^2 \ge 1 \implies x_ix_j \ge 1$. And so if $x_i \ge x_j$, $\frac {1+x_i^2}{x_i}-\frac {1+x_j^2}{x_j}=\frac{(x_ix_j-1)(x_i-x_j)}{x_ix_j} \ge 0$. Assume $\{x_i\}$ is increasing, using Chebyshev, we have $$n\sum\frac{x_i^2-n+1}{x_i} \ge (\sum\frac {x_i^2-n+1}{1+x_i^2})(\sum\frac {1+x_i^2}{x_i})=0.$$Thus $\sum\frac{x_i^2-n+1}{x_i} \ge 0 \implies \sum (x_i-\frac {n-1}{x_i}) \ge 0 \implies \sum x_i \ge (n-1)\sum\frac {1}{x_i}$. $\implies\frac{\sum x_i}{\sum \frac{1}{x_i}} \ge n-1$, the minimum is achieved when $x_1=x_2=...=x_n= \sqrt {n-1}$.

VicKmath7 wrote: Given are positive reals $x_1, x_2,..., x_n$ such that $\sum\frac {1}{1+x_i^2}=1$. Find the minimal value of the expression $\frac{\sum x_i}{\sum \frac{1}{x_i}}$ and find when it is achieved. It was here

Indeed. It's just a little "painted"