Let us denote by $A_1, \dots, A_N$ all the 700-element subsets of $\{1, 2, \dots, 1400\}$. Let $N = \binom{1400}{700}$. Now, choose 1400 polynomials $P_1, \dots, P_{1400}$ of the following form:
$$P_i(x) = 1 + 10^{10} \prod_{1 \le j \le N,\ j \in A_i} (x - j).$$
We now claim that if Alireza chooses these 1400 polynomials, he shall win the game. Assume that Amin chooses 700 polynomials $P_{i_1}, \dots, P_{i_{700}}$. Let us denote by $A_l$ the set $\{i_1, \dots, i_{700}\}$. Then, for all $i \in A_l$ we have $P_i(l) = 1$ hence $\Omega(P_i(l)) = 1$. Otherwise, $P_i(l) = 1 + 10^{10} K_i$ then $|P_i(l)| > 2$ and hence $\omega(P_i(l)) \ge 2 > 1$.

Hence, for $n = l$;
$$1 = \max_{i \in A_l} (\Omega(P_i(l))) < \min_{i \notin A_l} (\omega(P_i(l))).$$
Thus, by choosing such polynomials, Alireza shall win.