Consider the set $M=\{1,2,...,n\},n\in\mathbb N$. Find the smallest positive integer $k$ with the following property: In every $k$-element subset $S$ of $M$ there exist two elements, one of which divides the other one.

1. **Define the problem and the set \( M \):**
   We are given the set \( M = \{1, 2, \ldots, n\} \) where \( n \in \mathbb{N} \). We need to find the smallest positive integer \( k \) such that in every \( k \)-element subset \( S \) of \( M \), there exist two elements \( a \) and \( b \) such that \( a \) divides \( b \).

2. **Consider a specific subset \( S \):**
   Let's consider the subset \( S = \{n, n-1, \ldots, n-(k-1)\} \). We need to show that there are two elements \( a \) and \( b \) in \( S \) such that \( a \mid b \).

3. **Establish the condition for divisibility:**
   If \( a \mid b \), then \( a \leq \frac{b}{2} \). This implies that \( a \) must be at most half of \( b \).

4. **Derive the inequality:**
   On the other hand, we have \( n-k < a \leq \frac{b}{2} \leq n \). This gives us the inequality:
   \[
   n - k < \frac{n}{2}
   \]
   Simplifying this, we get:
   \[
   k > \frac{n}{2}
   \]

5. **Prove that \( k > \frac{n}{2} \) is sufficient:**
   Suppose there is a subset \( S \) of \( M \) with \( k \) elements that does not satisfy the condition. Consider subsets of the form \( S_{2k+1} = \{x = (2k+1) \cdot 2^l \mid l \in \mathbb{N}, x \leq n \} \) where \( 2k < n \).

6. **Analyze the intersection of \( S \) and \( S_{2k+1} \):**
   It is clear that \( |S_{2k+1} \cap S| \leq 1 \) because if there were two elements in the intersection, one would be a multiple of the other. The union \( \cup_{k=0}^{\frac{n}{2}} S_{2k+1} \) covers the set \( M \).

7. **Count the number of sets \( S_{2k+1} \):**
   The number of such sets \( S_{2k+1} \) is at most \( \frac{n}{2} \). Therefore, the size of \( S \) is less than or equal to the number of these sets, which is \( \frac{n}{2} \).

8. **Reach a contradiction:**
   If \( |S| \leq \frac{n}{2} \), then \( S \) cannot be a \( k \)-element subset with \( k > \frac{n}{2} \). This contradicts our assumption that \( S \) does not satisfy the condition.

Thus, the smallest positive integer \( k \) such that in every \( k \)-element subset \( S \) of \( M \), there exist two elements \( a \) and \( b \) such that \( a \mid b \) is \( k = \left\lceil \frac{n}{2} \right\rceil + 1 \).

The final answer is \( \boxed{ k = \left\lceil \frac{n}{2} \right\rceil + 1 } \).