Find the biggest natural number $m$ that has the following property: among any five 500-element subsets of $\{ 1,2,\dots, 1000\}$ there exist two sets, whose intersection contains at least $m$ numbers.
1. **Define the problem and the set construction:**
   We need to find the largest natural number \( m \) such that among any five 500-element subsets of \(\{1, 2, \dots, 1000\}\), there exist two sets whose intersection contains at least \( m \) numbers.

2. **Construct specific sets:**
   Consider the sets \( S_i = \{100(i-1) + 1, 100(i-1) + 2, \ldots, 100i\} \) for \( 1 \le i \le 10 \). These sets partition \(\{1, 2, \dots, 1000\}\) into 10 disjoint subsets, each containing 100 elements.

3. **Define the subsets \( A_i \):**
   Construct the following five 500-element subsets:
   \[
   \begin{aligned}
   A_1 &= S_1 \cup S_2 \cup S_3 \cup S_4 \cup S_5, \\
   A_2 &= S_1 \cup S_2 \cup S_6 \cup S_7 \cup S_8, \\
   A_3 &= S_3 \cup S_4 \cup S_6 \cup S_7 \cup S_9, \\
   A_4 &= S_1 \cup S_3 \cup S_8 \cup S_9 \cup S_{10}, \\
   A_5 &= S_2 \cup S_5 \cup S_6 \cup S_9 \cup S_{10}.
   \end{aligned}
   \]
   Each \( A_i \) is a union of 5 different \( S_i \) sets, ensuring each \( A_i \) has 500 elements.

4. **Calculate intersections:**
   We need to show that among any five such sets, there exist two sets whose intersection contains at least 200 elements. Consider the sum of the sizes of all pairwise intersections:
   \[
   \sum_{i < j} |A_i \cap A_j|
   \]
   By double-counting, this sum is equal to:
   \[
   \sum_{k=1}^{1000} \binom{x_k}{2}
   \]
   where \( x_k \) denotes the number of sets \( A_j \) containing the element \( k \).

5. **Apply discrete convexity:**
   Since \( x_1 + x_2 + \cdots + x_{1000} = 2500 \) and the \( x_i \) are integers, by the convexity of the function \( f(x) = \binom{x}{2} \), the sum is minimized when the \( x_i \) are as equal as possible. Given the constraint, the minimum sum is:
   \[
   500 \binom{3}{2} + 500 \binom{2}{2} = 500 \cdot 3 + 500 \cdot 1 = 2000
   \]

6. **Apply the Pigeonhole Principle:**
   There are \(\binom{5}{2} = 10\) terms in the sum \(\sum_{i < j} |A_i \cap A_j|\). By the Pigeonhole Principle, at least one of these terms must be at least:
   \[
   \frac{2000}{10} = 200
   \]
   Therefore, \( m \ge 200 \).

7. **Conclusion:**
   Since we have constructed sets achieving \( m = 200 \) and shown that \( m \ge 200 \), the largest such \( m \) is indeed 200.

The final answer is \( \boxed{200} \).