Let $p$ and $p+2$ be twin primes above 2013. For any positive integer $n$, we have $n$ and $n+p+2$ in different sets and also $n+2$ and $n+p+2$ in different sets; hence $n$ and $n+2$ are in the same set. Hence all even integers are in the same of $A$ and $B$, and all odd integers are in the same of $A$ and $B$. The case that one of $A$ or $B$ is empty is contradictive. The case that $A$ has all even and $B$ has all odd integers (or symmetric) is okay.

Complementary sequences?

We only need any two distinct odd primes $p,q$ that satisfy such condition. First note that if WLOG $p+r \in A$ for any nonnegative integer $r$, then $2p+r\in B, 3p+r\in A, 4p+r\in B,\dots$; in other words, $np+r$ and $mp+r$ are in the same subset if and only if $n\equiv m\mod{2}$. Now assume there exist two consecutive integers $\alpha$ and $\alpha+1$ in the same subset. Take your two distinct odd primes $p,q$, and by Bezout there exist integers $x,x’$ such that $$px+qx’=1$$Clearly $x$ and $x’$ are both nonzero and of different sign and parity; assume WLOG $x’$ is negative and let $y=-x’$. Then we have $px=qy+1\Rightarrow px+\alpha=qy+(\alpha+1)$. Since one of $x,y$ is odd and the other is even, one of $px+\alpha, qy+(\alpha+1)$ is in the same subset as $\alpha$ and $\alpha+1$, while the other one isn’t. This is absurd, since they are both the same number. It follows that no two elements in $A$ or $B$ are consecutive, which forces them to be the set of even and odd positive integers, in some order. This choice for $A$ and $B$ clearly works, since whenever the difference between any two positive integers is either $p$ or $q$ (and thus odd), they must be of different parity, hence they belong to different subsets. We are done. As a remark, we need at least two odd primes because if only one satisfies the condition, we get $p$ disjoint “strands” of numbers equal to each of the $p$ residues mod $p$, each totally independent from the other. In each of these strands, you can separate the elements into even and odd ones, where all evens and all odds belong to $A$ and $B$ in some order. So we have 2 choices for each strand, and the answer in this case is $2^p$ (unless there is a similar argument to prove you can’t have elements of different parity within the same subset, which I suspect there is). One could also generalize this to partition $\mathbb{Z}$ into $n$ disjoint subsets $\{ A_1,A_2,\dots,A_n\}$ such that $n$ distinct (odd) primes $p_1, p_2, \dots, p_n$ satisfy a similar condition; albeit unless it allows for a mod $n$ “rule” to be formulated similarly to the mod 2 rule we used earlier, it could be a bit challenging. On the former case though, one could try applying the same Bezout argument for each pair of primes $p_i, p_j$ and arriving to a nice contradiction, so as to prove the only such partition is into the $n$ subsets of positive integers with the same mod $n$ residue.

Same problem as https://artofproblemsolving.com/community/c6h146494p829199

ANSWER:One of $A,B$ is the set of all natural numbers.And the other one is set of all odd natural numbers.It indeed, satisfies the problem condition.We will prove that there is no other set. Lemma:Suppose $N$ is in $A$.Then for all $k\ge 1$ and prime $p\ge 2020$ we have all $N+(2k+1)p$ are in $B$ and $N+ 2kp$ are in $A$.Same thing will happen for an element in set $B$ proof: $N$ is in $A$.So $N+p$ is in $B$ which in turn implies that $N+2p$ is in $A$ and so on.$\blacksquare$ Now suppose 2 is in $A$. FTSOC, assume that an odd number m is in $A$.Take a prime $p\ge m+2020$. By dirichlet theorem, there is a $x$ such that $m-2+2xp$ is a prime greater than $2020^{2020}$.[As $gcd(m-2,2p)=1$]By the lemma $m+2xp$ is in $A$.But then $2, m+2xp$ are in $A$ and their difference is a prime greater than $2020^{2020}$. Contradiction.So no odd number is in $A$. So $1$ is in $B$.Again FTSOC, assume a even number $2n$ is in $B$.Take prime $p\ge 2n+2020$.By the lemma,all $2n+2kp$ are in $B$for all $k\ge 1$. By dirichlet theorem, there is $x$ such that $2n-1+2xp$ is a prime greater than $2020^{2020}$[As $gcd(2n-1,2p=1)=1$] But then $1,2n +2xp$ are in $B$ and their difference $2n-1+2xp$ is a prime greater than $2020^{2020}$.Contradiction.Hence no even number is in $B$. So it is clear that $A$ contains all even numbers.And $B$ contains all odd numbers.$\blacksquare$

Why are all sols so complicated? We show the only possibilities are separating the positive integers in odd and even numbers. Claim: $A$ cannot contain two consecutive numbers. This obviously implies the answer. Suppose so: Take $p,q$ big primes and $A,A+1$ two numbers in $A$. Then notice that $A+p \in B$, $A+2p \in A$ and so by induction $A+2lp \in A$ and similarly $A+(2k+1)q+1 \in B$, doing the same with $q$ and $A+1$. But by Bezout, there are $a,b$ such that $ap-bq=1$. Notice one of them is odd and the other is even. If $a$ is even, we have an absurd by making both the expressions equal setting $2l=a,2k+1=b$. If $a$ is odd, then just change $p,q$ in the argument above: $(2k+1)p+A \in B$ and $2lq+A+1 \in A$, and thus we can make them equal. $\blacksquare$