The counts of parity-odd vs even late-growing permutations in the group $\mathbb{S}_{2p-1}$ differ by $(-1)^p$

A permutation $σ\in\mathbb{S}_n$ of objects $\{1,\ldots,n\}$ is called \emph{late-growing} if its partial sums never exceed the respective partial sums of the `average' permutation (with the arithmetic mean $n^{-1}\cdot\sum_{i=1}^n i = (n+1)/2$ of $1,\ldots,n$ on average at each entry $σ(i)$): $\sum_{i=1}^k σ(i) \leqslant k\cdot (n+1)/2$ for all positive $k\leqslant n$. We prove that for $n=2p-1$ odd, the set $Φ_p\subseteq \mathbb{S}_{2p-1}$ of late-growing permutations is partitioned into parity-even vs odd elements, $Φ_p = Φ_p^{+} \sqcup Φ_p^{-}$, in an almost symmetric way: $| Φ_p^{+} | - | Φ_p^{-} | = (-1)^{p+1}$. The proof is constructive: by viewing the permutations $σ\inΦ_p$ as constrained walks on the integral lattice $\mathbb{Z}$, we describe two filters that pair all but one element in $Φ_p^{+}$ versus $Φ_p^{-}$; the remaining walk $σ^*$ with monotonically growing step amplitudes determines the dominant parity subset.