$A$ and $B$ wrote two positive integers respectively in a black board. The product of the numbers written by $A$ was twice as the sum of the numbers written by $B$ and the product of the numbers written by $B$ was twice as the sum of the numbers written by $A$ and the sum of the numbers written by $A$ is more than or equal to that of $B$. Find all possible sum written by $B$.
Note: Four numbers written are not always distinct.
Given the problem, we need to find all possible sums written by \( B \). Let's denote the numbers written by \( A \) as \( a \) and \( b \), and the numbers written by \( B \) as \( c \) and \( d \).

We are given the following conditions:
1. \( ab = 2(c + d) \)
2. \( cd = 2(a + b) \)
3. \( a + b \geq c + d \)

We start by manipulating the given equations to find a relationship between \( a, b, c, \) and \( d \).

1. From \( ab = 2(c + d) \) and \( cd = 2(a + b) \), we can write:
   \[
   ab + cd = 2(c + d) + 2(a + b)
   \]
   Rearranging, we get:
   \[
   ab + cd - 2a - 2b - 2c - 2d = 0
   \]

2. We can rewrite the above equation as:
   \[
   (a-2)(b-2) + (c-2)(d-2) = 8
   \]
   This is because:
   \[
   ab - 2a - 2b + 4 + cd - 2c - 2d + 4 = 8
   \]

3. Since \( (a-2)(b-2) \) and \( (c-2)(d-2) \) are even (as \( ab \) and \( cd \) are even), we need to check the cases where these products are even and sum to 8:
   \[
   ((a-2)(b-2), (c-2)(d-2)) = (0, 8), (2, 6), (4, 4)
   \]

4. We now check each case:

   - **Case 1: \( (a-2)(b-2) = 0 \) and \( (c-2)(d-2) = 8 \)**
     - If \( (a-2)(b-2) = 0 \), then either \( a = 2 \) or \( b = 2 \).
     - Without loss of generality, let \( a = 2 \). Then \( b \) can be any integer.
     - For \( (c-2)(d-2) = 8 \), possible pairs \((c, d)\) are \((3, 10)\) and \((4, 6)\).

   - **Case 2: \( (a-2)(b-2) = 2 \) and \( (c-2)(d-2) = 6 \)**
     - Possible pairs for \((a, b)\) are \((3, 4)\) and \((4, 3)\).
     - Possible pairs for \((c, d)\) are \((3, 6)\) and \((6, 3)\).

   - **Case 3: \( (a-2)(b-2) = 4 \) and \( (c-2)(d-2) = 4 \)**
     - Possible pairs for \((a, b)\) are \((4, 4)\).
     - Possible pairs for \((c, d)\) are \((4, 4)\).

5. We now verify each pair to ensure they satisfy the original conditions and \( a + b \geq c + d \):

   - For \((a, b) = (2, 13)\) and \((c, d) = (3, 10)\):
     \[
     ab = 2 \times 13 = 26, \quad 2(c + d) = 2(3 + 10) = 26
     \]
     \[
     cd = 3 \times 10 = 30, \quad 2(a + b) = 2(2 + 13) = 30
     \]
     \[
     a + b = 15 \geq c + d = 13
     \]

   - For \((a, b) = (2, 10)\) and \((c, d) = (4, 6)\):
     \[
     ab = 2 \times 10 = 20, \quad 2(c + d) = 2(4 + 6) = 20
     \]
     \[
     cd = 4 \times 6 = 24, \quad 2(a + b) = 2(2 + 10) = 24
     \]
     \[
     a + b = 12 \geq c + d = 10
     \]

   - For \((a, b) = (3, 6)\) and \((c, d) = (3, 6)\):
     \[
     ab = 3 \times 6 = 18, \quad 2(c + d) = 2(3 + 6) = 18
     \]
     \[
     cd = 3 \times 6 = 18, \quad 2(a + b) = 2(3 + 6) = 18
     \]
     \[
     a + b = 9 = c + d = 9
     \]

   - For \((a, b) = (4, 4)\) and \((c, d) = (4, 4)\):
     \[
     ab = 4 \times 4 = 16, \quad 2(c + d) = 2(4 + 4) = 16
     \]
     \[
     cd = 4 \times 4 = 16, \quad 2(a + b) = 2(4 + 4) = 16
     \]
     \[
     a + b = 8 = c + d = 8
     \]

Thus, the possible values of \( c + d \) are \( 13, 10, 9, \) and \( 8 \).

The final answer is \( \boxed{13, 10, 9, 8} \).