1. This should be easy, and the answer is $1011$. 2. Assume without loss of generality that $M_{P(x)}<1011$ and $P(0)$ is non-negative, otherwise just replace $P$ by $-P$. Notice that $P\in\mathbb{Z}[x]$ so $a-b\mid P(a)-P(b)$ for all integers $a,b$ and $|P(\alpha)|\le1010$ for $\alpha\in\{0,1,...,2021\}$. We have $|P(0)|\le 1011$. Write $P(\alpha)=P(0)+k\alpha$ for some integer $k$. If $|k|\ge2$ and $\alpha\ge1011$ then $$|P(0)+k\alpha|\ge|k|\alpha-|P(0)|\ge2022-1010>1011,$$which is not impossible. Consider $\alpha\ge1011$ then $|k|\le1$. Furthermore, $P(\alpha)$ cannot be $P(0)+\alpha$ since it follows from $P(0)\ge0$ that $P(0)+\alpha\ge\alpha>1010$. Now on, $\alpha$ should denote any integer in the set $\{1011,...,2021\}$ and $P(\alpha)$ can only be $P(0)$ or $P(0)-\alpha$. If there is some $\alpha$ such that $P(\alpha)=P(0)$, write $P(x)=x(x-\alpha)g(x)+P(0)$. For $x\in\{3,...,\alpha-3\}$ it's trivial to see that $|x(x-\alpha)|\ge3\alpha-9$, on the other hand $x(x-\alpha)$ divides $P(x)-P(0)$ and we must have $|P(x)|\le1010$, so if there exists some $\beta\in\{3,...,\alpha-3\}$ such that $g(\beta)$ is non-zero, it follows that $$|P(\beta)|\ge|x(x-\alpha)|-|P(0)|\ge3\alpha-1019>1010,$$which is obsolete. So $g(3)=...=g(\alpha-3)=0$, and now rewrite $P(x)-P(0)$ is divisible by $x(x-3)(x-4)...(x-\alpha+3)(x-\alpha)$. If $\deg P<\alpha-4$ then it quickly follows from the Lagrange's interpolation theorem we have $P(x)\equiv P(0)$, which is also obsolete. So we may write $P$ as $$P(x)=x(x-3)...(x-\alpha+3)(x-\alpha)h(x)+P(0),$$and continue the similar trick we arrive at $$P(x)=x(x-1)...(x-2021)+P(0),$$but this $P$ clearly does not satisfy $M_{P(x)}<1011$. So we may hereafter assume $P(\alpha)=P(0)-\alpha$ for all $\alpha$. Notice that $P(x)=P(x-1)-1$ for $x\in\{1011,...,2020\}$ then we have the following formula for $P$, which is $$P(x)=P(x-1)-1+\prod_{\alpha=1011}^{2020}(x-\alpha)p(x),$$and just like the above case we have $p(0)=p(1)=...=p(1010)=0$, and finally we have $P(x)=x(x-1)...(x-2021)$, which also does not satisfy. This concludes the proof of the problem.