Lemma. If $1 \le k_1 < k_2 < \dots < k_t \le n-1$ be natural numbers, then
$$\forall x \in \mathbb{R} : x^{2n} + \sum_{i=1}^{t} (x^{2k_i+1} + x^{2k_i}) + 1 > 0.$$
*Proof.* If $x > 0$, it's trivial. So let's assume that $x < 0$. If $-1 \le x < 0$, we write
$$x^{2n} + \sum_{i=1}^{t} (x^{2k_i+1} + x^{2k_i}) + 1$$
as
$$x^{2n} + 1 + (x+1) \sum_{i=1}^{t} x^{2k_i}.$$
Since $x + 1 \ge 0$ and $\sum_{i=1}^{t} x^{2k_i} \ge 0$, we have
$$x^{2n} + 1 + (x + 1) \sum_{i=1}^{t} x^{2k_i} > 0.$$
If $x < -1$, then $x + 1 < 0$ and $x^{2k+1} < x < 1$. So, $x^{2n+1}(x^{2k+1} + 1) > 0$. Therefore, we can re-write
$$x^{1398} + \sum_{i=1}^{t} (x^{2k_i+1} + x^{2k_i}) + 1$$
as
$$(x^{1398} - (2k_t + 1) + 1) x^{2k_t + 1} + \sum_{i=1}^{t-1} x^{2k_1+1} (x^{2k_{i+1}} - (2k_i + 1)) + 2^{2k_1} + 1$$
where $1 \le k_1 \le k_2 \le \dots \le \frac{1396}{2}$. So, it is positive. $\square$

Back to the problem. When Roozbeh chooses $x^k$, if $k$ is an even number, Keyvan should choose $x^k$ as well. But, if $k$ is an odd number, and if coefficient of $x^k$ is odd, Keyvan should choose $x^{k-1}$ and if coefficient of $x^k$ is even, Keyvan should play $x^{k+1}$. Obviously after Keyvan's move, the polynomial will be in the following form:
$$x^{1398} \sum_{i=1}^{m} (x^{s_i} + x^{s_i-1})^2 + 2 \sum_{i=1}^{n} x^{2t_i} + \sum_{i=1}^{l} (x^{2k_i+1} + x^{2k_i}) + 1.$$
Where
$$1 \le s_1 \le s_2 \le \dots \le s_m \le \frac{1398}{2},$$
$$0 \le t_1 \le t_2 \le \dots \le t_n \le \frac{1398}{2},$$
$$1 \le k_1 < k_2 < \dots < k_l \le 1398.$$
According to the lemma we proved, the polynomial should be always positive. $\blacksquare$