Vietnamese Mathematical Olympiad

Since the coefficients of $P(x)$ are non-negative, the root $x = p/q$ must be negative. Without loss of generality, we assume that $p < 0, q > 0$ then $|p| + n|q| = nq - p$. Let $R(x) = Q(x) - P(x)$, then $R(x)$ is an integer polynomial where $x = p/q$ is a root. This implies that
$$R(x) = (qx - p)T(x)$$
where $T(x)$ is a polynomial with rational coefficients. Since $\text{gcd}(p, q) = 1$, by using Gauss lemma for the product of two primitive polynomials, we get $T(x) \in \mathbb{Z}[x]$. This implies that $qn - p \mid R(n)$ for all $n = 1, 2, \dots, 2021$.

Hence, to finish the proof, we just need to show that $R(n) > 0$ for all $n = 1, 2, \dots, 2021$. Since $x = 2022$ is a root of $R(x)$, we also can write
$$R(x) = (x - 2022)H(x)$$
in which $H(x) = a_m x^m + a_{m-1} x^{m-1} + \dots + a_1 x + a_0$ is an polynomial with integer coefficients. Note that $n - 2022 < 0$ for all $n = 1, 2, \dots, 2021$, so it suffices to show that $H(n) < 0$ for all $n = 1, 2, \dots, 2021$.

By expanding the product in the right hand side, we can find out that the coefficient of $x^i$ is $a_{i-1} - 2022a_i$ for $1 \le i \le m$ and the constant is $-2022a_0$. On the other hand, since the coefficients of $P(x)$ are not larger than $2021$ then the coefficients of $R(x)$ are not less than $-2021$. From this, $a_0 \le 0$. Suppose that there exists some coefficients of $H(x)$ are positive, denote by $\ell > 0$, the smallest value of such indices. Thus, $a_\ell > 0$ and $a_{\ell-1} \le 0$ which implies that
$$a_{\ell-1} - 2022a_{\ell} \le 0 - 2022 = -2022.$$
This contradiction show that all coefficients of $H(x)$ is non-positive. But they can not all zero, otherwise $Q(x) \equiv P(x)$ but $Q(x)$ has at least one coefficient larger than $2021$. In conclusion, $H(n) < 0$ for all $n = 1, 2, \dots, 2021$. Therefore,
$$Q(n) - P(n) = R(n) \ge qn - p = |q|n + |p|,$$
for $n = 1, 2, \dots, 2021$. $\square$