Translating a homological/representation theoretic result into elementary things, I obtained the following (in case I made no mistake):


Let $n \geq 4$ and $w >3$ and let $w$ be an unit in $\mathbb{Z}/\mathbb{Z}n$.


Let $r:=2 \inf \{ s \geq 0 | sw+1 \equiv 0 $ mod $n \}$.


Let (\*) denote the following:
Given $a \in \mathbb{Z}/\mathbb{Z}n$ and $b \in \{1,...,w-1 \}$ with $a+\frac{1+(-1)^l}{2} b + w [\frac{l+1}{2}] \neq 0 $ mod $n$ for all $ l=1,2,...,r$, then $a \equiv 1-w$ and $b=w-1$. ([-] denotes the floor function: [2]=2, [3/2]=1)


Then $(\*)$ holds iff $w$ divides $n+1$.
Is there a simple and elementary interpretation/reason why this holds?



Elementary interpretation of a homological result
Denote $t=r/2=\min \{ s > 0 | sw+1 \equiv 0 \pmod n\}$. Our pair $(a,b)$ should satisfy $wm\ne -a;-(a+b)\pmod n$ for all $m=1,\dots,t$ (consider separately $l=2m$, $l=2m-1$). Note that for $-a=b=w-1$ this is always the case. Indeed, clearly $wm\ne 0=-(a+b)$, and $wm+a=w(m-1)+1\equiv w(m-t-1)$ also is not divisible by $n$.


Now assume that $w$ divides $n+1$, we should prove that there are no other pairs $(a,b)$ satisfying above property. Let $(a,b)$ be such a pair. Note that $(n+1)/w\equiv 1/w\equiv -t\pmod n$, thus $k:=(n+1)/w=n-t$. We have $m;m+b/w\ne -a/w$ for all $m=1,\dots,n-k$. It means that two segments $[1,n-k]$ and $b/w+[1,n-k]$ cover not all possible remainders modulo $n$. But $b/w\equiv bk\in \{k,2k,\dots,(w-1)k=n+1-k\}$. It is easily seen that unless $b=w-1$, the remainder of $bk$ lies between $k$ and $n-k$ and two segments cover all remainders; if $b=w-1$ there is unique not covered remainder $n-k+1=b/w$, so the pair $(a,b)$ is unique.


Now assume that $w$ does not divide $n+1$. Let us try to find another pair $(a,b)$, and look at pairs for which $a=-b$. Then we should find $b\in \{1,2,\dots,w-2\}$ such that $wm\ne -a=b\pmod n$ for all $m=1,\dots,t$. Choose $u\in \{2,\dots,w-1\}$ such that $n+u$ is divisible by $w$, write $n+u=qw$. Denote $k=n-t$, then $kw-1$ is divisible by $n$ and $kw-1\geqslant 2n$. It follows that $1<q<k$. Note that $(t+q)w\equiv u-1\in \{1,2,\dots,w-2\}$, but $t+q\in\{t+1,\dots,n-1\}$, so $t+q\notin \{1,2,\dots,t\}$. Therefore $wm\ne u-1$ for $m=1,\dots,t$ and we may take $b=u-1$. 


