We claim that Ali has a winning strategy for even $n$. If $n$ is even, Ali does the last move, assuming that he has to choose $a_k$. Choose a rational number $q$ which is not a root of $Q(x) = \sum_{i \neq k} a_i x^i$. Now Ali can choose $a_k = \frac{-Q(q)}{q^k}$ and win the game.

If $n$ is odd we show that Amin has a winning strategy. It is enough to prove the following lemma.

Lemma 1. *Assume that all of the coefficients are chosen but $a_k$, then one can choose $a_k$ in a way that $P(x)$ has no rational roots.*

Assume firstly that $k \neq 0, n$, and $d$ is the common denominator of $a_i$. Then for any natural number $t$, if $\frac{a}{b}$ is a root of $\sum_{i \neq k} a_i x^i + t x^k$, then $a \mid k a_0$, $b \mid k a_n$. So there are finitely many possibilities and we can choose $a_k$ in a way that avoids these possibilities.

If $i=0$, consider the polynomial $Q(x) = \sum_{i=1}^{n} a_i x_i$. It is enough to prove that there is a natural number $n$ so that $Q(x) = -a$ does not have a rational root. If $a$ is a natural number, as before, the denominator of any possible root is a divisor of $k a_0$. Now choose $x_0$ big enough such that for $|y| > x_0$, $|Q(y + \frac{1}{k a_0})| > |Q(y)| + 1$.

Now if $Q(y) = n$, then for every natural number $b$, $Q(y + \frac{b}{k a_0}) > n + 1$. So the equation $Q(x) = n + 1$ can't have any rational solution.

By considering the polynomial $R(x) = \sum a_{n-i} x^i$ and noting that the roots of $R$ are the inverse of the roots of $P$, we see that the case of $i = 0$ is equivalent to the case of $i = d$. We are done. ■