62nd Ukrainian National Mathematical Olympiad

Answer: No, not necessarily.

As an example, consider the following two polynomials:

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

Alternative solution. Consider the following two polynomials: $$P(x)=(x^3-1{)}^N,Q(x)=(x^2+x+1{)}^N.$$ 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: $$\frac{b}{a}\ge \frac{3^N}{2^N(2N+1)}>2023.$$