Find all natural numbers $n$ for which each natural number written with $~$ $n-1$ $~$ 'ones' and one 'seven' is prime.
To solve the problem, we need to find all natural numbers \( n \) such that any number formed by \( n-1 \) ones and one seven is prime. Let's denote such a number by \( N \).

1. **Formulation of the Number \( N \)**:
   The number \( N \) can be written as:
   \[
   N = 10^k + 10^{k-1} + \cdots + 10^m + 7 \times 10^l + \cdots + 10^0
   \]
   where \( k, m, l \) are positions of the digits, and \( l \) is the position of the digit '7'. 

2. **Modulo 7 Analysis**:
   We will use modulo 7 to analyze the primality of \( N \). Note that:
   \[
   10 \equiv 3 \pmod{7}
   \]
   Therefore, powers of 10 modulo 7 cycle every 6 steps:
   \[
   10^1 \equiv 3 \pmod{7}, \quad 10^2 \equiv 2 \pmod{7}, \quad 10^3 \equiv 6 \pmod{7}, \quad 10^4 \equiv 4 \pmod{7}, \quad 10^5 \equiv 5 \pmod{7}, \quad 10^6 \equiv 1 \pmod{7}
   \]

3. **Sum of Ones**:
   The sum of \( n-1 \) ones can be written as:
   \[
   S = 1 + 10 + 10^2 + \cdots + 10^{n-2}
   \]
   Using the geometric series sum formula:
   \[
   S = \frac{10^{n-1} - 1}{9}
   \]

4. **Modulo 7 of the Sum**:
   We need to find the modulo 7 of \( S \):
   \[
   S \equiv \frac{10^{n-1} - 1}{9} \pmod{7}
   \]
   Since \( 10^{n-1} \equiv 3^{n-1} \pmod{7} \):
   \[
   S \equiv \frac{3^{n-1} - 1}{9} \pmod{7}
   \]

5. **Adding the Seven**:
   The number \( N \) is:
   \[
   N = S + 7 \times 10^l
   \]
   Since \( 10^l \equiv 3^l \pmod{7} \):
   \[
   N \equiv \frac{3^{n-1} - 1}{9} + 7 \times 3^l \pmod{7}
   \]

6. **Checking for Primality**:
   We need to check if \( N \) is prime for all \( n \). We will consider different values of \( n \) modulo 3:
   - If \( n \equiv 0 \pmod{3} \), then \( n = 3k \):
     \[
     3^{n-1} \equiv 3^{3k-1} \equiv 3^{3(k-1)+2} \equiv 3^2 \equiv 2 \pmod{7}
     \]
     \[
     S \equiv \frac{2 - 1}{9} \equiv \frac{1}{9} \equiv 1 \pmod{7}
     \]
     Adding the seven:
     \[
     N \equiv 1 + 7 \times 3^l \pmod{7}
     \]
     Since \( 3^l \) cycles through \( 3, 2, 6, 4, 5, 1 \), \( N \) will not be prime for large \( n \).

   - If \( n \equiv 1 \pmod{3} \), then \( n = 3k+1 \):
     \[
     3^{n-1} \equiv 3^{3k} \equiv 1 \pmod{7}
     \]
     \[
     S \equiv \frac{1 - 1}{9} \equiv 0 \pmod{7}
     \]
     Adding the seven:
     \[
     N \equiv 0 + 7 \times 3^l \pmod{7}
     \]
     This will not be prime for large \( n \).

   - If \( n \equiv 2 \pmod{3} \), then \( n = 3k+2 \):
     \[
     3^{n-1} \equiv 3^{3k+1} \equiv 3 \pmod{7}
     \]
     \[
     S \equiv \frac{3 - 1}{9} \equiv \frac{2}{9} \equiv 2 \pmod{7}
     \]
     Adding the seven:
     \[
     N \equiv 2 + 7 \times 3^l \pmod{7}
     \]
     This will not be prime for large \( n \).

Conclusion:
For \( n = 1 \) and \( n = 2 \), the numbers are 7 and 17, which are prime. For \( n \geq 3 \), the numbers are not prime.

The final answer is \( \boxed{ n = 1 } \) and \( n = 2 \).