Find all the pairs positive integers $(x, y)$ such that $\frac{1}{x}+\frac{1}{y}+\frac{1}{[x, y]}+\frac{1}{(x, y)}=\frac{1}{2}$ ,
where $(x, y)$ is the greatest common divisor of $x, y$ and $[x, y]$ is the least common multiple of $x, y$.
1. **Assume \( x \leq y \)**. Let \((x, y) = d\), \(x = ds\), and \(y = dt\). Therefore, \([x, y] = dst\). The given equation is:
   \[
   \frac{1}{x} + \frac{1}{y} + \frac{1}{[x, y]} + \frac{1}{(x, y)} = \frac{1}{2}
   \]
   Substituting \(x\), \(y\), \([x, y]\), and \((x, y)\) into the equation, we get:
   \[
   \frac{1}{ds} + \frac{1}{dt} + \frac{1}{dst} + \frac{1}{d} = \frac{1}{2}
   \]
   Simplifying, we have:
   \[
   \frac{t + s + 1}{dst} + \frac{1}{d} = \frac{1}{2}
   \]
   Multiplying through by \(dst\), we get:
   \[
   t + s + 1 + st = \frac{dst}{2}
   \]
   Dividing through by \(d\), we get:
   \[
   \frac{s + t + 1}{st} = \frac{d - 2}{2}
   \]

2. **Analyze the range of \(d\)**. Note that \(d \geq 3\) because if \(d < 3\), the left-hand side of the equation would be too large to equal \(\frac{1}{2}\).

3. **Consider \(d \geq 9\)**. If \(d \geq 9\), then:
   \[
   \frac{s + t + 1}{st} \geq \frac{7}{2}
   \]
   This implies:
   \[
   7st - 2s - 2t - 2 \leq 0
   \]
   Rearranging, we get:
   \[
   25 \leq (7s - 2)(7t - 2) \leq 18
   \]
   This is a contradiction, so \(3 \leq d \leq 8\).

4. **Solve for specific values of \(d\)**:
   - For \(d = 3\):
     \[
     \frac{s + t + 1}{st} = \frac{1}{2}
     \]
     Solving \((s + t + 1) = \frac{st}{2}\), we get:
     \[
     2(s + t + 1) = st \implies st - 2s - 2t - 2 = 0
     \]
     Factoring, we get:
     \[
     (s - 2)(t - 2) = 6
     \]
     Solutions: \((s, t) = (3, 8)\) and \((4, 5)\).

   - For \(d = 4\):
     \[
     \frac{s + t + 1}{st} = 1
     \]
     Solving \((s + t + 1) = st\), we get:
     \[
     st - s - t - 1 = 0
     \]
     Factoring, we get:
     \[
     (s - 1)(t - 1) = 2
     \]
     Solutions: \((s, t) = (2, 3)\).

   - For \(d = 5\):
     \[
     \frac{s + t + 1}{st} = \frac{3}{2}
     \]
     Solving \((s + t + 1) = \frac{3st}{2}\), we get:
     \[
     2(s + t + 1) = 3st \implies 3st - 2s - 2t - 2 = 0
     \]
     Factoring, we get:
     \[
     (3s - 2)(3t - 2) = 8
     \]
     Solutions: \((s, t) = (1, 4)\).

   - For \(d = 6\):
     \[
     \frac{s + t + 1}{st} = 2
     \]
     Solving \((s + t + 1) = 2st\), we get:
     \[
     2st - s - t - 1 = 0
     \]
     Factoring, we get:
     \[
     (2s - 1)(2t - 1) = 3
     \]
     Solutions: \((s, t) = (1, 2)\).

   - For \(d = 7\):
     \[
     \frac{s + t + 1}{st} = \frac{5}{2}
     \]
     Solving \((s + t + 1) = \frac{5st}{2}\), we get:
     \[
     5st - 2s - 2t - 2 = 0
     \]
     Factoring, we get:
     \[
     (5s - 2)(5t - 2) = 8
     \]
     No integer solutions.

   - For \(d = 8\):
     \[
     \frac{s + t + 1}{st} = 3
     \]
     Solving \((s + t + 1) = 3st\), we get:
     \[
     3st - s - t - 1 = 0
     \]
     Factoring, we get:
     \[
     (3s - 1)(3t - 1) = 2
     \]
     Solutions: \((s, t) = (1, 1)\).

5. **List all solutions for \((x, y)\)**:
   - For \(d = 3\), \((s, t) = (3, 8)\) and \((4, 5)\):
     \[
     (x, y) = (9, 24), (12, 15)
     \]
   - For \(d = 4\), \((s, t) = (2, 3)\):
     \[
     (x, y) = (8, 12)
     \]
   - For \(d = 5\), \((s, t) = (1, 4)\):
     \[
     (x, y) = (5, 20)
     \]
   - For \(d = 6\), \((s, t) = (1, 2)\):
     \[
     (x, y) = (6, 12)
     \]
   - For \(d = 8\), \((s, t) = (1, 1)\):
     \[
     (x, y) = (8, 8)
     \]

6. **Combine all solutions**:
   \[
   (x, y) = (5, 20), (6, 12), (8, 8), (8, 12), (9, 24), (12, 15)
   \]

The final answer is \( \boxed{ (5, 20), (6, 12), (8, 8), (8, 12), (9, 24), (12, 15) } \)