Let's have a go for this one ! Let $S = \sum x_i$ Summing all equations and using $S$, we obtain the equivalent system of equation : $x_i = S \times (n-1) - 1/x_i$ or ${x_i}^2 - S/(n-1) \times x_i + 1 = 0$ From this we can deduct that all $x_i$ are $a$ or $1/a$ for some $a$ to be found. There are $n$ $'a'$ and $(n-k)$ $'1/a'$, for some k and $S = k \times a + (n-k)/a$ Replacing and solving equation gives : $a = \pm {\dfrac{\sqrt{\left (k-1\right )\left (k+1 -n\right )}}{k-1}}$. Roots are real (and $\neq 0$) iff $k = n$, i.e. $a = \pm 1/\sqrt{n-1}$ Solutions are then : $x_1 = \pm \sqrt{n-1}$, $x_j = \pm 1/\sqrt{n-1}$ and all permutations (with same $\pm$).

subtract any 2 of the equations and get either x(i)=1/x(j)or x(i)=x(j) let any x(1)=x(2)=...=x(k)=1/x(k+1)=....=1/x(n)=a (this is the only possibility) then a=(k-1)/a +(n-k)a so a^2=(k-1)/(1+k-n) this is positive so k>=n but obviously k<=n so k=n all the numbers are equal so, a=(n-1)/a we are through

Let \[a=\sum_{i=1}^n\frac{1}{x_i}.\] Obviosly all $x_i$ satisfyes equations $x_i+\frac{1}{x_i}=a$ or $x_i=\frac{a\pm\sqrt{a^2-4}}{2}\to |a|\ge 2.$ Let for $k$ of $x_i$ holds $x_i=\frac{a+\sqrt{a^2-4}}{2}$ and $n-k$ $x_i=\frac{a-\sqrt{a^2-4}}{2}$, then $2a=na+(n-2k)\sqrt{a^2-4}$ or $(2-n)^2a^2=(n-2k)^2(a^2-4)\to [(n-2k)^2-(n-2)^2]a^2=4(n-2k)^2$. If $n=1$, then $x_1=0$. If $n=2$ $x_1=x_2=\pm 1.$ If $n>2$, then $k=0$ $a^2=\frac{n^2}{n-1}.$