The answer is $2$. First of all, observe that Igor can ensure that the final polynomial has at most $2$ distinct integer roots, by simply setting $a_0 = 1$ in the first move. Now the final roots can only divide $1$, so either $-1$ or $1$. On the other hand, Egor can always ensure $x^2-1$ divides the final polynomial. He achieves this by dividing the coefficients into pairs by pairing $(a_{4k}, a_{4k+2});(a_{4k+1}, a_{4k+3})$ for all $0 \leq k \leq 24$. Now each time Igor sets a coefficient of some pair to a value, say $v$, Egor responds by setting the value of the other coefficient to $-v$. This ensures that the final polynomial is divisible by $x^2-1$ and hence has at least $2$ distinct roots, $\pm 1$ so we are done.