Determine all the sets of six consecutive positive integers such that the product of some two  of  them . added to the product of some other two  of  them is equal to the product of the remaining two numbers.
1. Let the six consecutive positive integers be \(a, a+1, a+2, a+3, a+4, a+5\).
2. We need to find sets of these integers such that the product of some two of them, added to the product of some other two of them, is equal to the product of the remaining two numbers.
3. We start by considering the inequality \(a(a+3) + (a+1)(a+2) \leq (a+4)(a+5)\).

   \[
   a(a+3) + (a+1)(a+2) \leq (a+4)(a+5)
   \]

4. Expanding and simplifying the inequality:

   \[
   a^2 + 3a + a^2 + 3a + 2 \leq a^2 + 9a + 20
   \]

   \[
   2a^2 + 6a + 2 \leq a^2 + 9a + 20
   \]

   \[
   a^2 - 3a - 18 \leq 0
   \]

5. Solving the quadratic inequality \(a^2 - 3a - 18 \leq 0\):

   \[
   a = \frac{3 \pm \sqrt{9 + 72}}{2} = \frac{3 \pm 9}{2}
   \]

   \[
   a = 6 \quad \text{or} \quad a = -3
   \]

   Since \(a\) must be a positive integer, we have \(a \leq 6\).

6. We now check each case for \(a = 1, 2, 3, 4, 5, 6\):

   - **Case \(a = 1\)**:
     \[
     1, 2, 3, 4, 5, 6
     \]
     \[
     1 \cdot 4 + 2 \cdot 3 = 4 + 6 = 10 \quad \text{and} \quad 5 \cdot 6 = 30
     \]
     This does not satisfy the condition.

   - **Case \(a = 2\)**:
     \[
     2, 3, 4, 5, 6, 7
     \]
     \[
     2 \cdot 5 + 3 \cdot 4 = 10 + 12 = 22 \quad \text{and} \quad 6 \cdot 7 = 42
     \]
     This does not satisfy the condition.

   - **Case \(a = 3\)**:
     \[
     3, 4, 5, 6, 7, 8
     \]
     \[
     3 \cdot 6 + 4 \cdot 5 = 18 + 20 = 38 \quad \text{and} \quad 7 \cdot 8 = 56
     \]
     This does not satisfy the condition.

   - **Case \(a = 4\)**:
     \[
     4, 5, 6, 7, 8, 9
     \]
     \[
     4 \cdot 7 + 5 \cdot 6 = 28 + 30 = 58 \quad \text{and} \quad 8 \cdot 9 = 72
     \]
     This does not satisfy the condition.

   - **Case \(a = 5\)**:
     \[
     5, 6, 7, 8, 9, 10
     \]
     \[
     5 \cdot 8 + 6 \cdot 7 = 40 + 42 = 82 \quad \text{and} \quad 9 \cdot 10 = 90
     \]
     This does not satisfy the condition.

   - **Case \(a = 6\)**:
     \[
     6, 7, 8, 9, 10, 11
     \]
     \[
     6 \cdot 9 + 7 \cdot 8 = 54 + 56 = 110 \quad \text{and} \quad 10 \cdot 11 = 110
     \]
     This satisfies the condition.

Conclusion:
The only set of six consecutive positive integers that satisfies the given condition is \(\{6, 7, 8, 9, 10, 11\}\).

The final answer is \( \boxed{ \{6, 7, 8, 9, 10, 11\} } \).