Find all natural numbers $k$ which can be represented as the sum of two relatively prime numbers not equal to $1$.
1. **Understanding the problem**: We need to find all natural numbers \( k \) that can be expressed as the sum of two relatively prime numbers, both greater than 1.

2. **Using Bertrand's Postulate**: Bertrand's Postulate states that for any integer \( n > 3 \), there is always at least one prime number \( p \) such that \( n < p < 2n \). For our problem, we need to find a prime \( p \) in the interval \( \left(\frac{k}{2}, k-1\right) \) for \( k > 6 \).

3. **Applying Bertrand's Postulate**: For \( k > 6 \), there exists a prime \( p \) such that \( \frac{k}{2} < p < k-1 \). This means \( k - p \) is also a natural number greater than 1. Since \( p \) is prime and \( k - p \) is less than \( p \), \( k - p \) and \( p \) are relatively prime.

4. **Verification for \( k \leq 6 \)**:
   - \( k = 2 \): The only pairs are (1,1), which are not greater than 1.
   - \( k = 3 \): The only pairs are (1,2) and (2,1), which are not both greater than 1.
   - \( k = 4 \): The only pairs are (1,3) and (3,1), which are not both greater than 1.
   - \( k = 5 \): The pairs are (2,3) and (3,2). Both pairs are greater than 1 and relatively prime.
   - \( k = 6 \): The pairs are (2,4), (4,2), and (3,3). None of these pairs are relatively prime.

5. **Conclusion**: From the above steps, we see that the only \( k \leq 6 \) that satisfies the condition is \( k = 5 \). For \( k > 6 \), Bertrand's Postulate guarantees that such a prime \( p \) exists, making \( k \) representable as the sum of two relatively prime numbers greater than 1.

The final answer is \( \boxed{ k = 5 } \) and all \( k \geq 7 \).