Find all the pairs $(n, m)$ of positive integers which fulfil simultaneously the conditions:
i) the number $n$ is composite;
ii) if the numbers $d_1, d_2, ..., d_k, k \in N^*$ are all the proper divisors of $n$, then the numbers $d_1 + 1, d_2 + 1, . . . , d_k + 1$ are all the proper divisors of $m$.
To solve the problem, we need to find all pairs \((n, m)\) of positive integers that satisfy the given conditions. Let's break down the problem step by step.

1. **Condition i: \(n\) is composite.**
   - A composite number is a positive integer greater than 1 that is not prime, meaning it has divisors other than 1 and itself.

2. **Condition ii: If \(d_1, d_2, \ldots, d_k\) are all the proper divisors of \(n\), then \(d_1 + 1, d_2 + 1, \ldots, d_k + 1\) are all the proper divisors of \(m\).**
   - Proper divisors of \(n\) are all divisors of \(n\) except \(n\) itself.

Let's denote the proper divisors of \(n\) as \(d_1, d_2, \ldots, d_k\) where \(1 = d_1 < d_2 < \cdots < d_k < n\). According to the problem, the numbers \(d_1 + 1, d_2 + 1, \ldots, d_k + 1\) must be the proper divisors of \(m\).

3. **Assume \(n\) is composite and list its proper divisors.**
   - For example, if \(n = 6\), the proper divisors are \(1, 2, 3\).

4. **Transform the proper divisors of \(n\) by adding 1 to each.**
   - For \(n = 6\), the transformed divisors are \(2, 3, 4\).

5. **Determine \(m\) such that these transformed divisors are the proper divisors of \(m\).**
   - For \(n = 6\), the proper divisors of \(m\) should be \(2, 3, 4\).

6. **Check if \(m = n + 1\) satisfies the conditions.**
   - For \(n = 6\), \(m = 7\). The proper divisors of 7 are \(1\), which does not match \(2, 3, 4\).

7. **Generalize the approach to find a pattern.**
   - Let's consider \(n = p_1 \cdot p_2\) where \(p_1\) and \(p_2\) are prime numbers. The proper divisors are \(1, p_1, p_2\).
   - The transformed divisors are \(2, p_1 + 1, p_2 + 1\).

8. **Check if \(m = p_1 \cdot p_2 + 1\) satisfies the conditions.**
   - For \(n = 6\), \(m = 7\) does not work.
   - For \(n = 4\), the proper divisors are \(1, 2\). The transformed divisors are \(2, 3\). If \(m = 5\), the proper divisors of 5 are \(1\), which does not match \(2, 3\).

9. **Re-evaluate the conditions and possible values of \(n\).**
   - Consider \(n = 8\). The proper divisors are \(1, 2, 4\). The transformed divisors are \(2, 3, 5\). If \(m = 9\), the proper divisors of 9 are \(1, 3\), which does not match \(2, 3, 5\).

10. **Conclude that no such pairs \((n, m)\) exist.**
    - After testing several composite numbers and their transformations, we find that no \(m\) satisfies the condition that the transformed divisors of \(n\) are the proper divisors of \(m\).

The final answer is \(\boxed{\text{No such pairs } (n, m) \text{ exist}}\).