We have two distinct positive integers $a,b$ with $a|b$. Each of $a,b$ consists of $2n$ decimal digits. The first $n$ digits of $a$ are identical to the last $n$ digits of $b$, and vice versa. Determine $a,b$.
1. Let \( a = x \cdot 10^n + y \) and \( b = y \cdot 10^n + x \), where \( x \) and \( y \) are \( n \)-digit numbers. This setup ensures that the first \( n \) digits of \( a \) are the same as the last \( n \) digits of \( b \), and vice versa.

2. Since \( a \mid b \), we can express \( b \) as \( b = ka \) for some integer \( k \). Substituting the expressions for \( a \) and \( b \), we get:
   \[
   y \cdot 10^n + x = k(x \cdot 10^n + y)
   \]

3. Rearrange the equation to isolate terms involving \( x \) and \( y \):
   \[
   y \cdot 10^n + x = kx \cdot 10^n + ky
   \]
   \[
   y \cdot 10^n - ky = kx \cdot 10^n - x
   \]
   \[
   y(10^n - k) = x(k \cdot 10^n - 1)
   \]

4. For \( a \) to divide \( b \), the remainder when \( b \) is divided by \( a \) must be zero. This implies:
   \[
   x(k \cdot 10^n - 1) = y(10^n - k)
   \]

5. To satisfy this equation, we need \( x \) and \( y \) such that \( x \mid y \) and \( y \mid x \). This implies \( x^2 = y \) because \( x \) and \( y \) must have the same number of digits.

6. Since \( x \) and \( y \) are \( n \)-digit numbers, and squaring any positive integer greater than 3 results in the number of digits increasing, we need to find \( x \) such that \( x^2 \) remains an \( n \)-digit number.

7. The only possible values for \( x \) and \( y \) that satisfy these conditions are:
   \[
   x = 24, \quad y = 24^2 = 576
   \]
   \[
   x = 39, \quad y = 39^2 = 1521
   \]

8. However, since \( x \) and \( y \) must have the same number of digits, we need to check if there are any other pairs that satisfy the conditions. The pairs \( (24, 576) \) and \( (39, 1521) \) do not satisfy the condition of having the same number of digits.

9. Therefore, the only valid pairs are \( (24, 42) \) and \( (39, 93) \), as these pairs satisfy the condition of having the same number of digits and the divisibility condition.

The final answer is \( \boxed{ (24, 42) } \) and \( (39, 93) \).