Let $A$ and $B$ be two sets such that $A \cup B$ is the set of the positive integers, and $A \cap B$ is the empty set. It is known that if two positive integers have a prime larger than $2013$ as their difference, then one of them is in $A$ and the other is in $B$. Find all the possibilities for the sets $A$ and $B$.
1. **Given Conditions:**
   - \( A \cup B \) is the set of all positive integers.
   - \( A \cap B \) is the empty set.
   - If two positive integers have a prime larger than 2013 as their difference, then one of them is in \( A \) and the other is in \( B \).

2. **Initial Assumption:**
   - Assume \( N \in A \).

3. **Lemma:**
   - Suppose \( N \in A \). Then for all \( k \ge 1 \) and prime \( p \ge 2020 \), we have:
     - \( N + (2k+1)p \in B \)
     - \( N + 2kp \in A \)
   - **Proof of Lemma:**
     - Since \( N \in A \), \( N + p \in B \) (because \( p \ge 2020 \)).
     - Then \( N + 2p \in A \) (since \( N + p \in B \)).
     - By induction, \( N + (2k+1)p \in B \) and \( N + 2kp \in A \) for all \( k \ge 1 \).

4. **Contradiction for Odd Numbers in \( A \):**
   - Suppose \( 2 \in A \).
   - Assume for contradiction that an odd number \( m \in A \).
   - Take a prime \( p \ge m + 2020 \).
   - By Dirichlet's theorem, there exists \( x \) such that \( m - 2 + 2xp \) is a prime greater than \( 2020^{2020} \) (since \( \gcd(m-2, 2p) = 1 \)).
   - By the lemma, \( m + 2xp \in A \).
   - But then \( 2 \) and \( m + 2xp \) are both in \( A \), and their difference is a prime greater than \( 2020^{2020} \), which contradicts the given condition.
   - Therefore, no odd number is in \( A \).

5. **Contradiction for Even Numbers in \( B \):**
   - Suppose \( 1 \in B \).
   - Assume for contradiction that an even number \( 2n \in B \).
   - Take a prime \( p \ge 2n + 2020 \).
   - By the lemma, all \( 2n + 2kp \in B \) for all \( k \ge 1 \).
   - By Dirichlet's theorem, there exists \( x \) such that \( 2n - 1 + 2xp \) is a prime greater than \( 2020^{2020} \) (since \( \gcd(2n-1, 2p) = 1 \)).
   - But then \( 1 \) and \( 2n + 2xp \) are both in \( B \), and their difference \( 2n - 1 + 2xp \) is a prime greater than \( 2020^{2020} \), which contradicts the given condition.
   - Therefore, no even number is in \( B \).

6. **Conclusion:**
   - \( A \) contains all even numbers.
   - \( B \) contains all odd numbers.

\(\blacksquare\)

The final answer is \( \boxed{ A } \) contains all even numbers and \( B \) contains all odd numbers.