VMO

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 $H(x)=a_m{x}^m+a_{m-1}{x}^{m-1}+\cdots +a_{1}x+a_0$ 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 $n=1,2,\dots ,2021$. $\square$