Consider $S=\{1, 2, 3, \cdots, 6n\}$, $n>1$. Find the largest $k$ such that the following statement is true: every subset $A$ of $S$ with $4n$ elements has at least $k$ pairs $(a,b)$, $a<b$ and $b$ is divisible by $a$.
1. **Definition**: Given an integer \( n = 2^a I \), where \( I \) is odd, we define \( I \) as the odd part of \( n \).

2. **Lemma 1**: \( k \ge n \).

   **Proof**: Consider the set \( S = \{1, 2, 3, \cdots, 6n\} \). The odd parts of the numbers in \( S \) are \( \{1, 3, 5, \cdots, 6n-1\} \). If two numbers \( a \) and \( b \) have the same odd part and \( a < b \), then \( a \mid b \).

   Let \( y_i \) be the number of elements in subset \( A \subset S \) with \( 4n \) elements that have the same odd part \( b_i \). Suppose \( \{1, 3, 5, \cdots, 6n-1\} = \{b_1, b_2, \cdots, b_{3n}\} \). Assume without loss of generality that:
   - \( y_1 = y_2 = \cdots = y_t = 0 \)
   - \( y_{t+1} = y_{t+2} = \cdots = y_l = 1 \)
   - \( y_{l+i} \ge 2 \) for all \( i \ge 1 \)

   Define \( \sum_{i=1}^{3n-l} y_{l+i} = S \). The number of pairs \( (a, b) \) such that \( a < b \) and \( a \mid b \) is at least:
   \[
   T = \sum_{i=1}^{3n-l} \binom{y_{l+i}}{2}
   \]
   Using the Cauchy-Schwarz inequality, we have:
   \[
   T \ge \frac{S^2 - (3n - l)S}{2(3n - l)}
   \]
   Since \( S + 1(l - t) + 0(t) = 4n \), we get \( S = 4n + t - l \). We also know \( S \ge 2(3n - l) \), thus:
   \[
   4n + t - l \ge 2(3n - l) \implies t + l \ge 2n
   \]
   Therefore:
   \[
   T \ge \frac{(4n + t - l)(n + t)}{2(3n - l)}
   \]
   We need to show that the right-hand side is at least \( n \):
   \[
   \frac{(4n + t - l)(n + t)}{2(3n - l)} \ge n
   \]
   This inequality holds if:
   \[
   (4n + t - l)(n + t) \ge 2n(3n - l)
   \]
   Simplifying, we get:
   \[
   4n^2 + 5nt + t^2 - ln - lt \ge 6n^2 - 2nl
   \]
   \[
   t(4n + t - l) + n(t + l) \ge 2n^2
   \]
   Since \( t + l \ge 2n \) and \( S, t \ge 0 \), the inequality holds. Therefore, \( k \ge T \ge n \).

3. **Lemma 2**: \( k \le n \).

   **Proof**: Consider the subset \( A = \{2n + 1, 2n + 2, \cdots, 6n\} \). It is easy to see that there are at most \( n \) pairs \( (a, b) \) such that \( a < b \) and \( a \mid b \).

4. Combining Lemma 1 and Lemma 2, we conclude that \( k = n \).

\(\blacksquare\)

The final answer is \( \boxed{ k = n } \)