Find all natural numbers $n$ for which every natural number whose decimal representation has $n-1$ digits $1$ and one digit $7$ is prime.
1. **Initial Observation**: We start by noting that \( n = 1 \) satisfies the requirement. For \( n = 1 \), the only number is 7, which is prime.

2. **General Form**: For \( n > 1 \), a number whose decimal representation has \( n-1 \) digits 1 and one digit 7 can be written as:
   \[
   \frac{10^n - 1}{9} + 6 \cdot 10^k
   \]
   where \( 0 \leq k \leq n-1 \).

3. **Divisibility by 7**: We need to find \( n \) such that \( \frac{10^n - 1}{9} + 6 \cdot 10^k \) is prime for every \( k = 0, 1, \dots, n-1 \). First, consider the term \( \frac{10^n - 1}{9} \). If \( \frac{10^n - 1}{9} \) is not divisible by 7, there exists some \( k_0 \) such that:
   \[
   \frac{10^n - 1}{9} + 6 \cdot 10^{k_0}
   \]
   is divisible by 7, and hence not prime.

4. **Divisibility by 7 Condition**: If \( \frac{10^n - 1}{9} \) is divisible by 7, then \( 10^n \equiv 1 \pmod{63} \). This implies that \( n \) must be a multiple of the order of 10 modulo 63. The order of 10 modulo 63 is 6, so \( n \) must be a multiple of 6.

5. **Divisibility by 3**: If \( n \) is a multiple of 6, then \( \frac{10^n - 1}{9} \) is divisible by 3. Therefore, for any \( k \), the number:
   \[
   \frac{10^n - 1}{9} + 6 \cdot 10^k
   \]
   is divisible by 3. Since the only prime number divisible by 3 is 3 itself, and \( \frac{10^n - 1}{9} + 6 \cdot 10^k \) is much larger than 3 for \( n > 1 \), it cannot be prime.

6. **Conclusion**: Therefore, there are no other \( n \) for which \( \frac{10^n - 1}{9} + 6 \cdot 10^k \) is prime for every \( k = 0, 1, \dots, n-1 \).

The final answer is \( \boxed{ n = 1 } \).