In other words,
$$\forall n,s\in\mathbb{N}, n\geq s$$
$$\begin{align}n\bmod{s}&=m_1\\(n+m_1)\bmod{s}&=m_2\\(n+m_2)\bmod{s}&=m_3\\\vdots\\(n+m_a)\bmod{s}&=0\end{align}$$
Is that guaranteed? I've only run a simple program and it seems it holds for values up to 500, but I'm unable to work it out mathematically.
By induction on $k$ one can prove that $kn$ gives the remainder $m_k$ modulo $s$ in your definitions. From $s|sn$ it follows that $m_s=0$.
There is a useful notation for such problems: we write $$a\equiv b(mod\ c)$$ when $a$ and $b$ give the same remainder modulo $c$, that is, $(a-b)|c$. The sign $\equiv$ looks similarly to identity sign, and that's because it is easy to show many similar properties like $a\equiv a;\ a\equiv b\Rightarrow b\equiv a;\ x\equiv y, y\equiv z\Rightarrow x\equiv z (mod\ c)$ $a\equiv b(mod\ c),x\equiv y(mod c)\Rightarrow a+x\equiv b+y(mod\ c),a-x\equiv b-y(mod\ c),ax\equiv by(mod c)$;
We have $\ {\rm mod}\ s\!:\,\ n_{\large 1}\equiv n,\,\ n_{\large k+1} \equiv n+n_{\large k}\ $ so $\ n_{\large k} \equiv kn\ $ by an easy induction.
Therefore $\ n_{\large k}\equiv 0\iff s\mid kn\iff s\mid(kn,ks) = k(n,s)\iff s/(n,s)\mid k$
i.e. $\,(n_{\large k}\bmod s) = 0\,$ iff $\,k\,$ is a multiple of $\, s/(n,s).\ $ Here $\, (n,s)\,$ denotes $\,\gcd(n,s).$