Let $ N $ be a positive integer. A set $ S \subset \{ 1, 2, \cdots, N \} $ is called [i]allowed[/i] if it does not contain three distinct elements $ a, b, c $ such that $ a $ divides $ b $ and $ b $ divides $c$. Determine the largest possible number of elements in an allowed set $ S $.
1. **Claim**: The largest possible number of elements in an allowed set \( S \) is \( N - \lfloor \frac{N}{4} \rfloor \).

2. **Construction**: Consider the set \( S \) consisting of all elements greater than \( \frac{N}{4} \). We need to show that this set is allowed, i.e., it does not contain three distinct elements \( a, b, c \) such that \( a \mid b \) and \( b \mid c \).

3. **Proof of Construction**:
    - Suppose \( a, b, c \in S \) with \( a < b < c \) and \( a \mid b \) and \( b \mid c \).
    - Since \( a \mid b \) and \( b \mid c \), it follows that \( c \geq 2b \geq 4a \).
    - However, since \( a > \frac{N}{4} \), we have \( 4a > N \), which contradicts the fact that \( c \leq N \).
    - Therefore, no such \( a, b, c \) exist, and the set \( S \) is allowed.

4. **Upper Bound**:
    - Every number \( n \) can be uniquely written as \( 2^l \cdot o \), where \( o \) is odd and \( l \) is a non-negative integer.
    - For an allowed set \( S \), there can be at most two elements with the same odd part \( o \) because if there were three, say \( 2^l \cdot o, 2^m \cdot o, 2^n \cdot o \) with \( l < m < n \), then \( 2^l \cdot o \mid 2^m \cdot o \) and \( 2^m \cdot o \mid 2^n \cdot o \), violating the allowed condition.

5. **Counting Elements**:
    - For \( \lfloor \frac{N}{2} \rfloor + 1 \leq o \leq N \) (where \( l = 0 \) by default), we can include at most all these numbers in our set. The total number of elements here is approximately \( \frac{N}{4} \).
    - For \( \lfloor \frac{N}{4} \rfloor + 1 \leq o \leq \lfloor \frac{N}{2} \rfloor \) (where \( l = 0 \) or \( l = 1 \)), we can include at most all these numbers in our set. The total number of elements here is approximately \( 2 \cdot \frac{N}{8} = \frac{N}{4} \).
    - For \( 1 \leq o \leq \lfloor \frac{N}{4} \rfloor \), we can include only two elements for each value of \( o \). The total number of elements here is approximately \( 2 \cdot \frac{N}{8} = \frac{N}{4} \).

6. **Combining Totals**:
    - Summing these contributions, we get approximately \( \frac{N}{4} + \frac{N}{4} + \frac{N}{4} = \frac{3N}{4} \).
    - Therefore, the largest possible number of elements in an allowed set \( S \) is \( N - \lfloor \frac{N}{4} \rfloor \).

\(\blacksquare\)

The final answer is \( \boxed{ N - \lfloor \frac{N}{4} \rfloor } \).