We claim that the second player has a winning strategy for $(m, n) = (p^{\alpha}, p^{\beta})$ where $p$ is a prime number and $\alpha, \beta \in \mathbb{Z}^{\ge 0}$. We call a pair $(m, n)$ a good pair if the second player has a winning strategy.

Lemma 1. If $(m, n)$ is a good pair and $d \mid m$ then $(d, n)$ is also a good pair.

*Proof.* Assume that the game starts with the sequence $a_1, a_2, ..., a_d$. Mohammad considers a second game that starts with the sequence $(\tilde{a}_1, \tilde{a}_2, ..., \tilde{a}_m)$ where $\tilde{a}_{\frac{mi}{d}} = a_i$ and $\tilde{a}_j = 0$, for $\frac{m}{d} \nmid j$. Now if Ali chooses $s$ in the first game, Mohammad assumes that Ali chooses $\frac{m}{d}s$ in the second game and follows his strategy and wins the second game. But the first game is just a restriction of the second game to the components $\frac{m}{d}i$ so he can also win the first game.

Lemma 2. $(m, nk)$ is a good pair if and only if $(m, n)$ and $(m, k)$ are also good pairs.

*Proof.* If $(m, nk)$ is a good pair then it is clear that $(m, n)$ and $(m, k)$ are good pairs. Now if $(m, n)$ and $(m, k)$ are good pairs, starting with a sequence $a_1, ..., a_m$ then using the aforementioned strategy for $(m, n)$, making all the components divisible by $n$. Considering the sequence $(\frac{a_i}{n})$, applying the preceding strategy for $(m, k)$ to ensure that $k \mid \frac{a_i}{n}$, by the very end all the terms are divisible by $nk$ and hence $nk$ is a good pair.

By these lemmas we see that if $(m, n)$ is a good pair and $p \mid m$, $q \mid n$ then $(p, q)$ is also a good pair. Now by the next lemma we would obtain $p = q$. Hence $(m, n)$ should be of the form $(p^{\alpha}, p^{\beta})$.

Lemma 3. If $(m, n)$ is a good pair and $m, n > 1$ then $\gcd(m, n) > 1$.

*Proof.* Assume that $\gcd(m, n) = 1$. Consider one step before the end of the game. The sequence $(a_i)$ must be constant; otherwise for each sequence $(b_i)$ there is some $j$ such that $a_i + b_j \not\equiv 0 \pmod{n}$ and the first player can choose $j - i$. In the second step before the last step, we should have
$$a_0 - a_m \equiv a_1 - a_0 = a_2 - a_1 \equiv \dots \equiv a_m - a_{m-1} \pmod{n},$$
otherwise Ali could choose a shift $s$ such that $(a_i + b_{i+s})$ is not constant. So we must have $m(a_1 - a_0) \equiv 0 \pmod{n}$, implying that $a_1 - a_0 \equiv 0 \pmod{n}$. So the sequence is constant. We then use induction to deduce that the sequence should be constant from the starting point. So, for a non-constant sequence the first player has a winning strategy.

It remains to prove that $(p^{\alpha}, p^{\beta})$ is a good pair. By the second lemma it suffices to prove that $(p^{\alpha}, p)$ is a good pair. We claim that if the second player chooses $b_i = -a_i$ at each step, he shall win. There are two ways to do this:

First proof. Let $T : (\mathbb{Z}/p\mathbb{Z})^m \to (\mathbb{Z}/p\mathbb{Z})^m$ denote a shift by 1. Then at each step the sequence changes to $(T^s - \text{id})(a_1, \dots, a_{p^\alpha})$. It suffices to prove that
$$\prod_{j=1}^{p^{\alpha}} (T^{s_j} - \text{id}) \equiv 0 \pmod{p}$$
but $T - \text{id} \mid T^{s_j} - \text{id}$. So it is enough to prove $(T - \text{id})^{p^\alpha} \equiv 0 \pmod{p}$ but
$$(T - \text{id})^{p^{\alpha}} \equiv T^{p^{\alpha}} - \text{id} \equiv 0 \pmod{p}.$$

Second proof.

Lemma 4. *There is an integer-valued polynomial $Q(x)$ such that $Q(i) \equiv a_i \pmod{p}$ for all $1 \le i \le p^\alpha$.*

*Proof.* Consider the polynomial $P_i(x) = \binom{x-i-1}{p^\alpha-1}$. One can see that $P_i(i) \equiv 1 \pmod{p}$ and for $j \ne i \pmod{p^\alpha}$, we have $P_i(j) \equiv 0 \pmod{p}$. Then we can choose
$$Q(x) = \sum_{i=1}^{p^{\alpha}} a_i P_i(x).$$
Now, in the next step we have $a_i \equiv Q(i) - Q(i+s) \pmod{p}$ but the degree of $Q(x) - Q(x+s)$ is less than $\deg(Q)$ so after $p^\alpha$ steps the second player wins.