Let $S$ be the set of all natural numbers with the property: the sum of the biggest three divisors of number $n$, different from $n$, is bigger than $n$. Determine the largest natural number $k$, which divides any number from $S$.
(A natural number is a positive integer)
1. **Understanding the Problem:**
   We need to find the largest natural number \( k \) that divides any number \( n \) in the set \( S \). The set \( S \) consists of natural numbers \( n \) such that the sum of the largest three divisors of \( n \), excluding \( n \) itself, is greater than \( n \).

2. **Analyzing the Condition:**
   Let \( d_1, d_2, d_3 \) be the largest three divisors of \( n \) excluding \( n \) itself. The condition given is:
   \[
   d_1 + d_2 + d_3 > n
   \]
   We need to find the largest \( k \) that divides every \( n \) in \( S \).

3. **Considering Smallest Divisors:**
   To satisfy the condition \( d_1 + d_2 + d_3 > n \), we start by considering the smallest possible divisors. The smallest divisors of any number \( n \) are typically 1, 2, 3, etc. However, since we are excluding \( n \) itself, we need to consider the next largest divisors.

4. **Testing with Small Numbers:**
   Let's test with small numbers to understand the pattern:
   - For \( n = 12 \):
     The divisors are \( 1, 2, 3, 4, 6, 12 \). The largest three divisors excluding 12 are \( 6, 4, 3 \).
     \[
     6 + 4 + 3 = 13 > 12
     \]
     So, 12 is in \( S \).
   - For \( n = 30 \):
     The divisors are \( 1, 2, 3, 5, 6, 10, 15, 30 \). The largest three divisors excluding 30 are \( 15, 10, 6 \).
     \[
     15 + 10 + 6 = 31 > 30
     \]
     So, 30 is in \( S \).

5. **Generalizing the Pattern:**
   From the examples, we see that the smallest divisors \( d_1 \) and \( d_2 \) must be 2 and 3 respectively to ensure the sum of the largest three divisors is greater than \( n \). This implies that \( n \) must be divisible by both 2 and 3.

6. **Finding the Largest \( k \):**
   Since \( n \) must be divisible by both 2 and 3, the largest \( k \) that divides any number in \( S \) is the least common multiple (LCM) of 2 and 3:
   \[
   \text{LCM}(2, 3) = 6
   \]

Conclusion:
\[
\boxed{6}
\]