Let $ S$ be the set of all odd positive integers less than $ 30m$ which are not multiples of $ 5$, where $ m$ is a given positive integer. Find the smallest positive integer $ k$ such that each $ k$-element subset of $ S$ contains two distinct numbers, one of which divides the other.
1. **Identify the set \( S \):**
   The set \( S \) consists of all odd positive integers less than \( 30m \) that are not multiples of 5. We need to count these numbers.

2. **Count the odd numbers less than \( 30m \):**
   The total number of positive integers less than \( 30m \) is \( 30m - 1 \). Half of these are odd, so there are:
   \[
   \left\lfloor \frac{30m - 1}{2} \right\rfloor + 1
   \]
   odd numbers less than \( 30m \).

3. **Count the odd multiples of 5 less than \( 30m \):**
   The odd multiples of 5 less than \( 30m \) are of the form \( 5(2k+1) \). We need to find the largest \( k \) such that \( 5(2k+1) < 30m \):
   \[
   5(2k+1) < 30m \implies 2k+1 < 6m \implies k < \frac{6m-1}{2}
   \]
   So, the number of odd multiples of 5 less than \( 30m \) is:
   \[
   \left\lfloor \frac{6m-1}{2} \right\rfloor
   \]

4. **Calculate the size of \( S \):**
   The size of \( S \) is the total number of odd numbers less than \( 30m \) minus the number of odd multiples of 5:
   \[
   |S| = \left\lfloor \frac{30m - 1}{2} \right\rfloor + 1 - \left\lfloor \frac{6m-1}{2} \right\rfloor
   \]

5. **Determine the smallest \( k \):**
   We need to find the smallest \( k \) such that any \( k \)-element subset of \( S \) contains two distinct numbers, one of which divides the other. This is related to the Erdős–Szekeres theorem, which states that in any sequence of more than \( n \) numbers, there is a subsequence of length \( n+1 \) that is either strictly increasing or strictly decreasing.

   For our problem, we need to ensure that any subset of size \( k \) contains two numbers where one divides the other. This is a problem of finding the largest subset of \( S \) where no number divides another. 

   The largest such subset is formed by the primes in \( S \). Therefore, \( k \) must be at least the number of primes in \( S \).

6. **Count the primes in \( S \):**
   The primes in \( S \) are the odd primes less than \( 30m \) that are not multiples of 5. The number of such primes can be approximated using the prime number theorem, but for exact counting, we need to list them.

   For small values of \( m \), we can list the primes directly. For larger values, we use the fact that the density of primes around \( n \) is approximately \( \frac{n}{\ln n} \).

   For simplicity, let's assume \( m \) is large enough that we can use the approximation:
   \[
   \pi(30m) \approx \frac{30m}{\ln(30m)}
   \]
   where \( \pi(x) \) is the number of primes less than \( x \).

   The number of primes not divisible by 5 is approximately:
   \[
   \pi(30m) - \pi(6m)
   \]

7. **Conclusion:**
   The smallest \( k \) is at least the number of primes in \( S \). For exact values, we need to count the primes directly, but the approximation gives us a good estimate.

The final answer is \( \boxed{ k \geq \pi(30m) - \pi(6m) } \).