Find all positive integers $A,B$ satisfying the following properties:
(i) $A$ and $B$ has same digit in decimal. 
(ii) $2\cdot A\cdot B=\overline{AB}$ (Here $\cdot$ denotes multiplication, $\overline{AB}$ denotes we write $A$ and $B$ in turn. For example, if $A=12,B=34$, then $\overline{AB}=1234$)
To find all positive integers \( A \) and \( B \) satisfying the given properties, we need to follow these steps:

1. **Express the given condition mathematically:**
   The problem states that \( 2 \cdot A \cdot B = \overline{AB} \). Here, \(\overline{AB}\) denotes the concatenation of \( A \) and \( B \). If \( A \) has \( n \) digits, then \(\overline{AB} = 10^n \cdot A + B \).

2. **Set up the equation:**
   \[
   2AB = 10^n A + B
   \]

3. **Rearrange the equation:**
   \[
   2AB - B = 10^n A
   \]
   \[
   B(2A - 1) = 10^n A
   \]

4. **Analyze the divisibility:**
   Since \( B \) must be a positive integer, \( 2A - 1 \) must divide \( 10^n A \). We also know that \( A \) and \( B \) have the same number of digits, so \( B \) can be expressed as \( kA \) where \( k \) is a single-digit integer.

5. **Substitute \( B = kA \) into the equation:**
   \[
   kA(2A - 1) = 10^n A
   \]
   \[
   k(2A - 1) = 10^n
   \]

6. **Determine possible values for \( k \):**
   Since \( k \) is a single-digit integer and must divide \( 10^n \), the possible values for \( k \) are \( 1, 2, 4, 5, 8 \). However, since \( 2A - 1 \) must be even (as \( 10^n \) is even), \( k \) must be even. Therefore, \( k \) can be \( 2, 4, 8 \).

7. **Check each case for \( k \):**

   - **Case 1: \( k = 2 \)**
     \[
     2(2A - 1) = 10^n
     \]
     \[
     4A - 2 = 10^n
     \]
     \[
     4A = 10^n + 2
     \]
     \[
     A = \frac{10^n + 2}{4}
     \]
     For \( n = 1 \):
     \[
     A = \frac{10 + 2}{4} = 3
     \]
     \[
     B = 2A = 6
     \]
     So, \( (A, B) = (3, 6) \).

   - **Case 2: \( k = 4 \)**
     \[
     4(2A - 1) = 10^n
     \]
     \[
     8A - 4 = 10^n
     \]
     \[
     8A = 10^n + 4
     \]
     \[
     A = \frac{10^n + 4}{8}
     \]
     For \( n = 2 \):
     \[
     A = \frac{100 + 4}{8} = 13
     \]
     \[
     B = 4A = 52
     \]
     So, \( (A, B) = (13, 52) \).

   - **Case 3: \( k = 8 \)**
     \[
     8(2A - 1) = 10^n
     \]
     \[
     16A - 8 = 10^n
     \]
     \[
     16A = 10^n + 8
     \]
     \[
     A = \frac{10^n + 8}{16}
     \]
     For \( n = 3 \):
     \[
     A = \frac{1000 + 8}{16} = 63
     \]
     \[
     B = 8A = 504
     \]
     However, \( B \) must have the same number of digits as \( A \), and \( 504 \) has more digits than \( 63 \). Therefore, this case does not provide a valid solution.

Conclusion:
The valid solutions are \( (A, B) = (3, 6) \) and \( (13, 52) \).

The final answer is \(\boxed{(3, 6), (13, 52)}\).