power of 2 involving floor function divides cyclic product if $S_n={a_1,a_2,a_3,...,a_{2n}}$} where $a_1,a_2,a_3,...,a_{2n}$ are all distinct integers.Denote by $T$ the product
$$T=\prod_{i,j\epsilon S_n,i<j}{(a_i-a_j})$$ Prove that $2^{(n^2-n+2[\frac{n}{3}])}\times \text{lcm}(1,3,5,...,(2n-1)) $ divides $T$.(where [] is the floor function)
I have tried many approaches.I tried using the fact that either number of odd or even integers$\ge(n+1)$ and since in $T$ every integer is paired with every other integer only once.Since even-even and odd-odd both give even numbers:$2^{\frac{n(n+1)}{2}}$ divides $T$.As for the $lcm(1,,3,5,....(2n-1))$ I have no Idea what to do.Please help.I am quite new to NT.Thank you.

Let me give a sketch of the solution.
Well, there is a following fact:

Proposition. Given $n$ distinct positive integers $a_1, a_2, \ldots, a_n$. Then,
  $$
\prod_{i>j}\frac{a_i-a_j}{i-j}\in \mathbb{Z}.
$$
  In other words, $A_n$ divides $\prod\limits_{i>j}(a_i-a_j)$, where $A_n=\prod\limits_{i>j}(i-j)$.

This fact can be proved by showing that number of multiplicands of the product $\prod\limits_{i>j}(a_i-a_j)$ which are divisible by $p^s$ for every prime $p$ and positive integer is greater or equal the number of multiplicands of the product $\prod\limits_{i>j}(i-j)$ which are divisible by $p^s$.
It's clear that in the statement of the proposition above constant $A_n$ is sharp (because we can simply consider $a_k=k$ for $k=\overline{1, n}$).
Thus, it's sufficient to prove that $A_{2n}$ is divisible by $2^{(n^2-n+2[\frac{n}{3}])}\times\text{lcm}(1,3,5,...,(2n-1))$. Since $A_{2n}$ is divisible by $1, 3, \ldots, 2n-1$ we need to show that 
$$
2^{n^2-n+2[\frac{n}{3}]}|A_{2n}.
$$
This can be done in the way which is described in the previous answer.
