The first key observation is that it suffices to consider the equation $$(x_1+x_2+\dots+x_n)\left(\frac{1}{x_1}+\frac{1}{x_2}+\dots+\frac{1}{x_n}\right)=4n^2$$As if $(x_1,x_2,\dots,x_n)$ satisfies this equation then there exists a $\lambda$ such that $(\lambda x_1,\lambda x_2,\dots,\lambda x_n)$ satisfies the original equation and $\frac{x_n}{x_1}$ remains the same. Let $f(i,j)$ denote $\frac{x_i}{x_j}+\frac{x_j}{x_i}$. The equation becomes $$\sum_{i<j} f(i,j)=4n^2-n$$ Claim: If $i<j<k$ then $f(i,j)+f(j,k)\leq 2+f(i,k)$ The function of $x_j$, $f(i,j)+f(j,k)$, is of the form $\frac{x_j}{a}+\frac{b}{x_j}$ so it is maximized at one of the endpoints, finishing the claim. We now claim that if $\frac{x_n}{x_1}<7+4\sqrt{3}$ then $$\sum_{1\leq i<j\leq n} f(i,j)<4n^2-n$$Notice we have that $f(i,j)<14$. We finish by induction by $2$ with the trivial base cases $n=1,2$. When going from $n\mapsto n+2$ we will use the induction hypothesis on $x_2,x_3,\dots,x_{n+1}$ $$\sum_{1\leq i<j\leq n+2} f(i,j)=f(1,n\;+\;2)+\sum_{2\leq i\leq n\;+\;1}(f(1,i)\;+\;f(i,n+2))\;+\;\sum_{2\leq i<j\leq n+1}f(i,j)<$$$$f(1,n\;+\;2)(n\;+\;1)\;+\;2n\;+\;4n^2-n<14(n\;+\;1)\;+\;2n\;+\;4n^2\;-\;n=4(n\;+\;2)^2\;-\;(n\;+\;2)$$