Let $n$ be a positive integer and $S_n$ be the symmetric group on $\{1,2,\ldots,n\}$. Let $\mathcal{A}_n$ be the anti-symmetrization operator on $\mathbb{Z}[x_1,x_2,\ldots,x_n]$ such that for any $f(x_1,x_2,\ldots,x_n)\in \mathbb{Z}[x_1,x_2,\ldots,x_n]$, $$\mathcal{A}_n(f)=\sum_{w\in S_n}\varepsilon(w)f(x_{w(1)},x_{w(2)},\ldots,x_{w(n)}),$$ where $\varepsilon(w)=(-1)^{l(w)}$ is the sign of $w$.

**My question:**

For any integer $n>1$ with $n(n-1)\equiv 0 \pmod 4$, is it true that we can always choose $\displaystyle \frac{n(n-1)}{4}$ different polynomials $$x_{i_k}+x_{j_k},\ 1\leq k\leq \frac{n(n-1)}{4}$$ from the $\displaystyle \frac{n(n-1)}{2}$ polynomials $$x_i+x_j,\ 1\leq i<j\leq n,$$ such that $$\mathcal{A}_n\left(\prod_{1\leq k\leq \frac{n(n-1)}{4}}(x_{i_k}+x_{j_k})^2\right)\neq0\ ?$$

For example, \begin{align*} n=4,\ \ &\mathcal{A}_4\left((x_1+x_2)^2(x_2+x_3)^2(x_3+x_4)^2\right)\neq0;\\ n=5,\ \ &\mathcal{A}_5\left((x_1+x_2)^2(x_2+x_3)^2(x_3+x_4)^2(x_4+x_5)^2(x_5+x_1)^2\right)\neq0. \end{align*}