Find all positive integers $n$ for which there exist two distinct numbers of $n$ digits, $\overline{a_1a_2\ldots a_n}$ and  $\overline{b_1b_2\ldots b_n}$, such that the number of $2n$ digits $\overline{a_1a_2\ldots a_nb_1b_2\ldots b_n}$ is divisible by $\overline{b_1b_2\ldots b_na_1a_2\ldots a_n}$.
1. Let \( A = \overline{a_1a_2 \ldots a_n} \) and \( B = \overline{b_1b_2 \ldots b_n} \). According to the given conditions, \( A \) and \( B \) are distinct \( n \)-digit positive integers. This implies:
   \[
   10^{n-1} \leq A, B < 10^n
   \]

2. The number \( \overline{a_1a_2 \ldots a_nb_1b_2 \ldots b_n} = 10^nA + B \) is divisible by \( \overline{b_1b_2 \ldots b_na_1a_2 \ldots a_n} = 10^nB + A \). Therefore, we have:
   \[
   10^nA + B = k(10^nB + A)
   \]
   for some positive integer \( k \).

3. Rearrange the equation:
   \[
   10^nA + B = k \cdot 10^nB + kA
   \]
   \[
   10^nA - kA = k \cdot 10^nB - B
   \]
   \[
   (10^n - k)A = (10^nk - 1)B
   \]

4. Let \( d = \gcd(10^n - k, 10^nk - 1) \). Then:
   \[
   \gcd\left(\frac{10^n - k}{d}, \frac{10^nk - 1}{d}\right) = 1
   \]
   This implies:
   \[
   \frac{10^nk - 1}{d} \mid A
   \]
   Hence:
   \[
   A \geq \frac{10^nk - 1}{d}
   \]

5. Since \( 10^{n-1} \leq A < 10^n \), we have:
   \[
   10^{n-1} \leq \frac{10^nk - 1}{d} < 10^n
   \]
   This implies:
   \[
   10^nk - 1 < d \cdot 10^n
   \]
   \[
   k < d
   \]
   Since \( k > 1 \), we have \( d \geq 2 \).

6. We also have:
   \[
   10^n \equiv k \pmod{d}
   \]
   \[
   0 \equiv 10^nk - 1 \equiv k^2 - 1 \pmod{d}
   \]
   \[
   d \mid (k-1)(k+1)
   \]

7. Since \( d > 1 \), \( d \) has a prime divisor \( p \). From \( d \mid (k-1)(k+1) \), we have \( p \mid k \pm 1 \). Therefore, \( p \leq k+1 \leq 10 \). Since \( p \not\in \{2, 5\} \) (as \( p \mid 10^nk - 1 \)), we have \( p \in \{3, 7\} \).

8. If \( p = 3 \), then \( 9 \mid 10^nk - 1 \) implies \( 9 \mid k - 1 \), which is impossible since \( 2 \leq k \leq 9 \).

9. Therefore, \( 7 \mid d \). By the Division Algorithm, there exist non-negative integers \( q \) and \( r < 6 \) such that \( n = 6q + r \). Hence:
   \[
   10^n = 10^{6q + r} = 10^r \cdot (10^6)^q \equiv 10^r \pmod{7}
   \]
   By Fermat's Little Theorem, \( 10^6 \equiv 1 \pmod{7} \). Thus:
   \[
   10^n \equiv 10^r \pmod{7}
   \]

10. Since \( 7 \mid \gcd(10^n - k, 10^nk - 1) = d \), we have \( 7 \mid 10^r - k \) and \( 7 \mid 10^rk - 1 \). Moreover, \( 7 \mid k \pm 1 \), yielding \( k \in \{6, 8\} \) since \( 2 \leq k \leq 9 \).

11. **Case 1:** \( k = 6 \)
    - Then \( 7 \mid 10^r - 6 \) and \( 7 \mid 6 \cdot 10^r - 1 \) iff \( r = 3 \).
    - Thus, when \( n \equiv 3 \pmod{6} \), the integers \( A = \frac{6 \cdot 10^n - 1}{7} \) and \( B = \frac{10^n - 6}{7} \) satisfy the conditions.

12. **Case 2:** \( k = 8 \)
    - Then \( 7 \mid 10^r - 8 \) and \( 7 \mid 8 \cdot 10^r - 1 \) iff \( r = 0 \).
    - Furthermore, \( d \mid 10^r - 8 \) and \( 10^r - 8 \equiv 1^r - 8 = -7 \equiv 2 \pmod{3} \) implies \( 3 \not\mid d \).
    - This combined with \( d \mid k^2 - 1 = 8^2 - 1 = 63 = 3^2 \cdot 7 \) gives us \( d \mid 7 \).
    - Consequently, \( d = 7 \), which is impossible since \( d \geq k = 8 \).

13. Hence, there are no integers \( A \) and \( B \) satisfying the conditions in this case.

Conclusion:
There exist two integers \( A \) and \( B \) satisfying the conditions if and only if \( n \equiv 3 \pmod{6} \).

The final answer is \( \boxed{ n \equiv 3 \pmod{6} } \)