Let us have $n$ ( $n>3$) balls with different rays. On each ball it is written an integer number. Determine the greatest natural number $d$ such that for any numbers written on the balls, we can always find at least 4 different ways to choose some balls with the sum of the numbers written on them divisible by $d$.
1. **Understanding the Problem:**
   We need to determine the greatest natural number \( d \) such that for any set of \( n \) balls (where \( n > 3 \)) with integers written on them, there are at least 4 different ways to choose some balls such that the sum of the numbers on the chosen balls is divisible by \( d \).

2. **Using the Given Lemma:**
   The lemma states that if we have \( n \) numbers \( a_1, a_2, \ldots, a_n \), we can always find a group of these numbers whose sum is divisible by \( n \). This is a direct application of the Pigeonhole Principle.

3. **Applying the Lemma:**
   Let \( S_i = \sum_{k=1}^i a_k \) for \( i = 1, 2, \ldots, n \). If any \( S_i \) is divisible by \( n \), we have found our group. If none of the \( S_i \) are divisible by \( n \), then by the Pigeonhole Principle, there must be at least two sums \( S_i \) and \( S_j \) (with \( i > j \)) such that \( S_i \equiv S_j \pmod{n} \). This implies \( S_i - S_j = a_{j+1} + a_{j+2} + \cdots + a_i \) is divisible by \( n \).

4. **Finding the Greatest \( d \):**
   We need to ensure that there are at least 4 different ways to choose subsets of balls such that their sums are divisible by \( d \). Since the lemma guarantees at least one such subset for \( n \), we need to extend this to ensure at least 4 such subsets.

5. **Considering \( d = \frac{n}{4} \):**
   If \( d = \frac{n}{4} \), then we need to find 4 different subsets whose sums are divisible by \( \frac{n}{4} \). This is feasible because:
   - For \( n \) numbers, there are \( 2^n \) possible subsets.
   - By the Pigeonhole Principle, if we consider the sums modulo \( \frac{n}{4} \), there are \( \frac{n}{4} \) possible remainders.
   - Since \( n > 3 \), \( \frac{n}{4} \) is a natural number, and there are enough subsets to ensure at least 4 different sums that are divisible by \( \frac{n}{4} \).

6. **Conclusion:**
   The greatest natural number \( d \) such that there are always at least 4 different ways to choose subsets of balls with sums divisible by \( d \) is \( \frac{n}{4} \).

The final answer is \( \boxed{\frac{n}{4}} \).