Find the least positive integer which is a multiple of $13$ and all its digits are the same.

[i](Adapted from Gazeta Matematică 1/1982, Florin Nicolăită)[/i]
1. We need to find the smallest positive integer that is a multiple of \(13\) and has all its digits the same. Let this number be represented as \(a \cdot \frac{10^{n+1} - 1}{9}\), where \(a\) is a digit from 1 to 9, and \(n+1\) is the number of digits.

2. Since \(a\) is a digit and we want the smallest such number, we can start by assuming \(a = 1\). This simplifies our problem to finding the smallest \(n\) such that:
   \[
   \frac{10^{n+1} - 1}{9} \equiv 0 \pmod{13}
   \]
   This is equivalent to:
   \[
   10^{n+1} - 1 \equiv 0 \pmod{13}
   \]
   or
   \[
   10^{n+1} \equiv 1 \pmod{13}
   \]

3. To solve \(10^{n+1} \equiv 1 \pmod{13}\), we need to find the order of 10 modulo 13. The order of an integer \(a\) modulo \(n\) is the smallest positive integer \(k\) such that \(a^k \equiv 1 \pmod{n}\).

4. We calculate the powers of 10 modulo 13:
   \[
   10^1 \equiv 10 \pmod{13}
   \]
   \[
   10^2 \equiv 100 \equiv 9 \pmod{13}
   \]
   \[
   10^3 \equiv 10 \cdot 9 \equiv 90 \equiv 12 \pmod{13}
   \]
   \[
   10^4 \equiv 10 \cdot 12 \equiv 120 \equiv 3 \pmod{13}
   \]
   \[
   10^5 \equiv 10 \cdot 3 \equiv 30 \equiv 4 \pmod{13}
   \]
   \[
   10^6 \equiv 10 \cdot 4 \equiv 40 \equiv 1 \pmod{13}
   \]

5. We see that \(10^6 \equiv 1 \pmod{13}\). Therefore, the order of 10 modulo 13 is 6. This means \(10^{n+1} \equiv 1 \pmod{13}\) when \(n+1\) is a multiple of 6.

6. The smallest \(n+1\) that is a multiple of 6 is 6 itself. Therefore, \(n = 5\).

7. The number with \(n+1 = 6\) digits, all of which are 1, is:
   \[
   111111
   \]

8. We verify that \(111111\) is a multiple of 13:
   \[
   111111 \div 13 = 8547
   \]
   Since 8547 is an integer, \(111111\) is indeed a multiple of 13.

The final answer is \(\boxed{111111}\).