Ukrainian National Mathematical Olympiad

Answer: No, not necessarily.

As an example, consider the following two polynomials:

$$Q(x) = 1 + 2x + 3x^2 + 4x^3 + \cdots + 2023x^{2022} + 2024x^{2023} + 2023x^{2024} + \cdots + x^{4046}, \quad R(x) = x - 1,$$
$$\text{then } P(x) = Q(x)(x-1) = -1 - x - x^2 - \cdots - x^{2023} + x^{2024} + x^{2025} + \cdots + x^{4047}.$$
Then $a = 1$ and $b = 2024$, with $b > 2023a$.

Alternative solution. Consider the following two polynomials:
$$P(x) = (x^3 - 1)^N, \quad Q(x) = (x^2 + x + 1)^N.$$
By the construction of $P(x) : Q(x)$, that is, the polynomial $R(x)$ exists. Since
$$P(x) = (x^3 - 1)^N = \sum_{j=0}^{N} C_N^j x^{3j} \Rightarrow a = C_N^j \text{ for some } j = 0, N \Rightarrow a = C_N^j < \sum_{j=0}^{N} C_N^j = (1+1)^N = 2^N.$$
The sum of the coefficients of the polynomial $Q(x)$ is equal to $Q(1) = (1 + 1 + 1)^N = 3^N$ and the number of coefficients is $2N + 1$, because it has degree $2N$. Therefore, according to Dirichlet's principle, there is a coefficient at least
$$\frac{3^N}{2N + 1} \Rightarrow b \ge \frac{3^N}{2N + 1}.$$
It is clear that with sufficiently large $N$ inequality will hold:
$$\frac{b}{a} \ge \frac{3^N}{2^N(2N + 1)} > 2023.$$