If we square elements in each set, we get $n$ is odd. But rest is not easy i think.

Easy to see that $n>2$ For $n=2$ we must have $a_1=a_2$ As the sets are equals, we have $\prod_1^na_i=\prod_1^n({a_i+\frac{(-1)^i}{a_i})}$ wich give us: $\prod_1^n\frac{1}{a_i^2}=\prod_1^n\frac{1}{{a_i}^2+{(-1)^i} }$ Now take $a_i=b_ix^\frac{1}{2}$ with $x>0$ $(\prod_1^n\frac{1}{b_i^2}).x^\frac{-n}{2}=\prod_1^n\frac{1}{{b_i}^2x+(-1)^i }=\sum_1^n\frac{c_i}{{b_i}^2x+(-1)^i}$ with ${c_i=\prod_{j=1 j<>i}^n}\frac{b_i}{(-1)^ib_j+(-1)^jb_i}$ wich exist as $b_i<>b_j$ Taking $\int_0^1dx$ in both sides gives: $\frac{2}{2-n}=(\prod_1^n{b_i}^2)(\sum_1^n\frac{c_i ln|{b_i}^2+(-1)^i|}{{b_i}^2})=A$ So $n=\frac{2(A-1)}{A}$ the only integer on this forme is $n=2$ Such sets don't exist

This is a nice problem I think. The answer is all the odd integers. Hint: Like Mathematicalx did, we know even numbers don't satisfy the condition. Finally, try to prove the set exists for every odd integers. I would give my proof if necessary.

AYMANE, there are so many errors in your write-up, it's maybe worth for you to try and identify at least a few ...

Give your's solution BSJL please!

Solution from Korea MOP Indeed $n$ must be odd integer larger than $1$. Also, the set does exist for such integers. Denote $f(x)=x+\frac{1}{x}$, $g(x)=x-\frac{1}{x}$. Let $h(x)=g(f(g(f( \cdots g(x) \cdots )-x$, where there are $k+1$ iterations of $g$ and $k$ iterations of $l$. It suffices to prove that $h(x)$ has a root. Note that $f(x),g(x)$ are strictly increasing for $x>1$. Since $f(x) \ge 2$ for all $x>0$, we have $g(f(x)) \ge g(2) = \frac{3}{2}$. Therefore, we have $h(\frac{3}{2})>0$. Now note that $g(f(x))<x \iff x> \frac{1+\sqrt{5}}{2}$. Let $x>f^k(\frac{1+\sqrt{5}}{2})$. $f^k(x)$ denotes $f$ iterated $k$ times. If for all $1 \le l \le k$ we have $g(f(\cdots g(x) \cdots ) \ge \frac{1+\sqrt{5}}{2}$, ($l+1$ iterations of $g$, $l$ iterations of $f$) We have $$g(f( \cdots g(x) \cdots ) \text{ } (k+1\text{ iterations of }g) \le g(f( \cdots g(x) \cdots ) \text{ } (k\text{ iterations of }g) \cdots \le g(x)=x-\frac{1}{x} < x $$which gives $h(x)<0$. If there exists there exists an $l$ such that $g(f( \cdots g(x) \cdots) < \frac{1+\sqrt{5}}{2}$, we have $$g(f( \cdots g(x) \cdots ) \text{ } (k+1\text{ iterations of }g) < g(f( \cdots f(\frac{1+\sqrt{5}}{2}) \cdots) (k-l\text{ iterations of }g) < f^{k-l} (\frac{1+\sqrt{5}}{2}) < f^k(\frac{1+\sqrt{5}}{2}) < x$$which gives $h(x)<0$. Now there must be an $x$ such that $h(x)=0$. Therefore, the answer is all odd natural numbers excluding $1$.