3 added 806 characters in body

I don't have an answer but here is a formulation which seems a bit more combinatorial.

Let's start defining some objects. Let $S_n$ be the set of all vectors which are a permutation of $\{1,2,\dots,n\}$, and let's add a partial order structure given by dominance (majorization), $(\sigma(1),\dots,\sigma(n))\succ (\tau(1,\dots,\tau(n)))$ iff $$\sum_{i=1}^k \sigma(i)\geq \sum_{i=1}^k \tau(i)$$ for all $1\le k\le n$. Now, when given $y_1,y_2,\cdots,y_n$, let us reorder them in non-inreasing order $y'1\geq\cdots\geq y'_n$, this clearly doesn't affect the problem. One observation is that if $\epsilon(\pi)\neq 0$ and $\pi'\succ \pi$ in $(S_n,\succ)$ then $\epsilon(\pi') \neq 0$. This implies that for every $\vec{x},\vec{y}$, the set of permutations $\sigma\in S_n$ for which $\epsilon(\sigma) \neq 0$ is an upper set, or actually an order ideal of $(S_n,\succ)$ since there is a maximal element.

Let's define $\chi(P)$ for an upper set in $(S_n,\succ)$ as $\sum_{\sigma\in P} \sgn operatorname{sgn} (\sigma)$, so that now the problem becomes

Show that $\chi(P)\le C^n \sqrt{n!}$ for all upper sets $P$.

We have $\chi{S_n}=0$ and $\chi$ of a chain is almost bounded (the signs alternate), which makes this result close to following from a Dilworth type result. Unfortunately the chains aren't long enough to partition an upper set in a "small" number of chains (i.e. there are large antichains). However $\chi$ seems very close to some sort of Mobius function for this poset, showing the similarity you mention in the post. Maybe there is an argument that can prove the inequality from here...

Edit: Let me also add that our poset's diagram seems to be the Permutohedron plus a few more edges between consecutive levels. In particular it seems to have (almost) the same grading. Just intuition tells me that the maximum should be achieved when $P$ is the upper half of the poset. That is $\chi$ is at most of the order of the alternating sum $\sum_{k=0}^{\frac{1}{2}\binom{n}{2}} (-1)^k a(k)$ where $a(k)$ is the number of permutations with $k$ inversions. Maybe this can be made rigorous?

Also, some googling reveals that once one puts a weight on each element of the poset, there is a problem of determining the ideal of maximum weight (sum of weights of its elements), so maybe some of that literature might be helpful.

Post Undeleted by Gjergji Zaimi
2 added 687 characters in body

I don't have an answer but here is a formulation which seems a bit more combinatorial.

Let's start defining some objects. Let $S_n$ be the set of all vectors which are a permutation of $\{1,2,\dots,n\}$, and let's add a partial order structure given by dominance (majorization), $(\sigma(1),\dots,\sigma(n))\succ (\tau(1,\dots,\tau(n)))$ iff $$\sum_{i=1}^k \sigma(i)\geq \sum_{i=1}^k \tau(i)$$ for all $1\le k\le n$. Now, when given $y_1,y_2,\cdots,y_n$, let us reorder them in non-inreasing order $y'1\geq\cdots\geq y'_n$, this clearly doesn't affect the problem. One observation is that if $\epsilon(\pi)\neq 0$ and $\pi'\succ \pi$ in $(S_n,\succ)$ then $\epsilon(\pi') \neq 0$. This implies that for every $\vec{x},\vec{y}$, the set of permutations $\sigma\in S_n$ for which $\epsilon(\sigma) \neq 0$ is an upper set, or actually an order ideal of $(S_n,\succ)$ since there is a maximal element.

Let's define $\chi(P)$ for an upper set in $(S_n,\succ)$ as $\sum_{\sigma\in P} \sgn (\sigma)$, so that now the problem becomes

Show that $\chi(P)\le C^n \sqrt{n!}$ for all upper sets $P$.

We have $\chi{S_n}=0$ and $\chi$ of a chain is almost bounded (the signs alternate), which makes this result close to following from a Dilworth type result. Unfortunately the chains aren't long enough to partition an upper set in a "small" number of chains (i.e. there are large antichains). However $\chi$ seems very close to some sort of Mobius function for this poset, showing the similarity you mention in the post. Maybe there is an argument that can prove the inequality from here...

Post Deleted by Gjergji Zaimi
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