There are two possible families of solutions: - $P(x)=x+d$, for some integer $d \geq-2022$. - $P(x)=-x+d$, for some integer $d \leq 2022$. Suppose $P$ satisfies the problem conditions. Clearly $P$ cannot be a constant polynomial. Notice that a polynomial $P$ satisfies the conditions if and only if $-P$ also satisfies them. Hence, we may assume the leading coefficient of $P$ is positive. Then, there exists positive integer $M$ such that $P(x)>0$ for $x \geq M$. Lemma 1. For any positive integer $n$, the integers $P(1), P(2), \ldots, P(n)$ leave pairwise distinct remainders upon division by $n$. Proof. Assume for contradiction that this is not the case. Then, for some $1 \leq y<z \leq n$, there exists $0 \leq r \leq n-1$ such that $P(y) \equiv P(z) \equiv r(\bmod n)$. Since $P(a n+b) \equiv P(b)(\bmod n)$ for all $a, b$ integers, we have $P(a n+y) \equiv P(a n+z) \equiv r(\bmod n)$ for any integer $a$. Let $A$ be a positive integer such that $A n \geq M$, and let $k$ be a positive integer such that $k>2 A+2021$. Each of the $2(k-A)$ integers $P(A n+y), P(A n+z), P((A+1) n+y), P((A+1) n+z), \ldots, P((k-1) n+y), P((k-1) n+z)$ leaves one of the $k$ remainders $$r, n+r, 2 n+r, \ldots,(k-1) n+r$$ upon division by $k n$. This implies that at least $2(k-A)-k=k-2 A$ (possibly overlapping) pairs leave the same remainder upon division by $k n$. Since $k-2 A>2021$ and all of the $2(k-A)$ integers are positive, we find more than 2021 pairs $a, b$ with $a<b \leq k n$ for which $|P(b)|-|P(a)|$ is divisible by $k n$ - hence, $P_{k n}>2021$, a contradiction. Next, we show that $P$ is linear. Assume that this is not the case, i.e., $\operatorname{deg} P \geq 2$. Then we can find a positive integer $k$ such that $P(k)-P(1) \geq k$. This means that among the integers $P(1), P(2), \ldots, P(P(k)-P(1))$, two of them, namely $P(k)$ and $P(1)$, leave the same remainder upon division by $P(k)-P(1)$ - contradicting the lemma (by taking $n=P(k)-P(1)$ ). Hence, $P$ must be linear. We can now write $P(x)=c x+d$ with $c>0$. We prove that $c=1$ by two ways. Solution 1 If $c \geq 2$, then $P(1)$ and $P(2)$ leave the same remainder upon division by $c$, contradicting the Lemma. Hence $c=1$. Solution 2 Suppose $c \geq 2$. Let $n$ be a positive integer such that $n>2 c M, n\left(1-\frac{3}{2 c}\right)>2022$ and $2 c \mid n$. Notice that for any positive integers $i$ such that $\frac{3 n}{2 c}+i<n, P\left(\frac{3 n}{2 c}+i\right)-P\left(\frac{n}{2 c}+i\right)=n$. Hence, $\left(\frac{n}{2 c}+i, \frac{3 n}{2 c}+i\right)$ satisfies the condition in the question for all positive integers $i$ such that $\frac{3 n}{2 c}+i<n$. Hence, $P_{n}>2021$, a contradiction. Then, $c=1$. If $d \leq-2023$, then there are at least 2022 pairs $a<b$ such that $P(a)=P(b)$, namely $(a, b)=$ $(1,-2 d-1),(2,-2 d-2), \ldots,(-d-1,-d+1)$. This implies that $d \geq-2022$. Finally, we verify that $P(x)=x+d$ satisfies the condition for any $d \geq-2022$. Fix a positive integer $n$. Note that $| P(b)|-| P(a)|<n$ for all positive integers $a<b \leq n$, so the only pairs $a, b$ for which $|P(b)|-|P(a)|$ could be divisible by $n$ are those for which $|P(a)|=|P(b)|$. When $d \geq-2022$, there are indeed at most 2021 such pairs.