The sets $A$ and $B$ are subsets of the positive integers. The sum of any two distinct elements of $A$ is an element of $B$. The quotient of any two distinct elements of $B$ (where we divide the largest by the smallest of the two) is an element of $A$. Determine the maximum number of elements in $A\cup B$.
1. **Assume \( |A| \geq 3 \)**:
   - Let \( x < y < z \) be the three smallest elements of \( A \).
   - Since the sum of any two distinct elements of \( A \) is in \( B \), we have:
     \[
     x + y, \quad x + z, \quad y + z \in B
     \]

2. **Consider the quotients**:
   - The quotient of any two distinct elements of \( B \) (where we divide the largest by the smallest) is an element of \( A \). Therefore:
     \[
     \frac{y+z}{x+y}, \quad \frac{y+z}{x+z}, \quad \frac{x+z}{x+y} \in A
     \]

3. **Analyze the inequalities**:
   - Since \( x < y < z \), we have:
     \[
     \frac{y+z}{x+y} < \frac{y+z}{x+z} < z
     \]
   - This implies:
     \[
     \frac{x+z}{x+y} < \frac{y+z}{x+y}
     \]

4. **Equate the quotients**:
   - Given that the quotients are elements of \( A \), we must have:
     \[
     \frac{x+z}{x+y} = x, \quad \frac{y+z}{x+z} = x, \quad \frac{y+z}{x+y} = y
     \]

5. **Solve the equations**:
   - From \( \frac{x+z}{x+y} = x \), we get:
     \[
     x + z = x^2 + xy \implies z = x^2 + xy - x
     \]
   - From \( \frac{y+z}{x+z} = x \), we get:
     \[
     y + z = x^2 + xz \implies z = x^2 + xz - y
     \]
   - From \( \frac{y+z}{x+y} = y \), we get:
     \[
     y + z = xy + y^2 \implies z = xy + y^2 - y
     \]

6. **Combine the equations**:
   - Equating the expressions for \( z \):
     \[
     x^2 + xy - x = xy + y^2 - y
     \]
   - Simplifying, we get:
     \[
     x^2 - x = y^2 - y \implies x(x-1) = y(y-1)
     \]

7. **Analyze the factors**:
   - Since \( x \) and \( y \) are distinct positive integers, the only solution to \( x(x-1) = y(y-1) \) is \( x = y \), which is a contradiction.

8. **Conclude \( |A| \leq 2 \)**:
   - Therefore, \( |A| \leq 2 \).

9. **Determine the maximum number of elements in \( A \cup B \)**:
   - If \( |A| = 2 \), let \( A = \{a, b\} \) with \( a < b \).
   - Then \( B \) must contain \( a+b \) and any other elements such that their quotients are in \( A \).
   - A possible configuration is \( A = \{2, 4\} \) and \( B = \{6, 12, 24\} \), giving \( |A \cup B| = 5 \).

The final answer is \( \boxed{5} \).