Determine the smallest natural number $n =>2$ with the property:
 For every positive integers $a_1, a_2,.  .  .  , a_n$  the product of all  differences $a_j-a_i$, 
  $1 <=i <j <=n$, is divisible by 2001.
To determine the smallest natural number \( n \geq 2 \) such that for every set of positive integers \( a_1, a_2, \ldots, a_n \), the product of all differences \( a_j - a_i \) (for \( 1 \leq i < j \leq n \)) is divisible by 2001, we need to consider the prime factorization of 2001 and the properties of differences.

1. **Prime Factorization of 2001:**
   \[
   2001 = 3 \times 23 \times 29
   \]

2. **Divisibility Condition:**
   We need the product of all differences \( a_j - a_i \) to be divisible by 2001. This means that the product must be divisible by each of the prime factors 3, 23, and 29.

3. **Pigeonhole Principle:**
   To ensure that the product of differences is divisible by each prime factor, we need to have at least one pair of \( a_i \) and \( a_j \) such that \( a_j - a_i \) is divisible by each of these primes.

4. **Minimum Number of Elements:**
   - For the product to be divisible by 3, we need at least 3 elements (since with 2 elements, the difference might not be divisible by 3).
   - For the product to be divisible by 23, we need at least 23 elements.
   - For the product to be divisible by 29, we need at least 29 elements.

5. **Combining the Conditions:**
   To ensure that the product is divisible by 2001, we need to satisfy the conditions for all prime factors simultaneously. Therefore, we need at least:
   \[
   n = 1 + \max(3, 23, 29) = 1 + 29 = 30
   \]

Thus, the smallest \( n \) such that the product of all differences \( a_j - a_i \) is always divisible by 2001 is \( n = 30 \).

The final answer is \( \boxed{30} \).