A positive integer $k\geqslant 3$ is called[i] fibby[/i] if there exists a positive integer $n$ and positive integers $d_1 < d_2  < \ldots < d_k$ with the following properties:  \\  $\bullet$ $d_{j+2}=d_{j+1}+d_j$ for every $j$ satisfying $1\leqslant j \leqslant k-2$, \\ $\bullet$ $d_1, d_2, \ldots, d_k$ are divisors of $n$, \\ $\bullet$ any other divisor of $n$ is either less than $d_1$ or greater than $d_k$.

 Find all fibby numbers. \\ \\ [i]Proposed by Ivan Novak.[/i]
1. **Base Cases:**
   - We first check the smallest values of \( k \) to see if they are fibby.
   - For \( k = 3 \):
     - Consider \( d_1 = 1 \), \( d_2 = 2 \), and \( d_3 = 3 \). These satisfy \( d_{j+2} = d_{j+1} + d_j \) for \( j = 1 \).
     - Let \( n = 6 \). The divisors of \( n \) are \( 1, 2, 3, 6 \). The divisors \( 1, 2, 3 \) are \( d_1, d_2, d_3 \), and the only other divisor \( 6 \) is greater than \( d_3 \).
     - Thus, \( k = 3 \) is fibby.
   - For \( k = 4 \):
     - Consider \( d_1 = 1 \), \( d_2 = 2 \), \( d_3 = 3 \), and \( d_4 = 5 \). These satisfy \( d_{j+2} = d_{j+1} + d_j \) for \( j = 1, 2 \).
     - Let \( n = 30 \). The divisors of \( n \) are \( 1, 2, 3, 5, 6, 10, 15, 30 \). The divisors \( 1, 2, 3, 5 \) are \( d_1, d_2, d_3, d_4 \), and the other divisors are either less than \( d_1 \) or greater than \( d_4 \).
     - Thus, \( k = 4 \) is fibby.

2. **General Case for \( k \geq 5 \):**
   - Assume \( k \geq 5 \) and suppose \( d_1 < d_2 < \ldots < d_k \) are such that \( d_{j+2} = d_{j+1} + d_j \) for \( 1 \leq j \leq k-2 \).
   - **Claim:** All \( d_i \) with \( i \geq 3 \) are odd.
   - **Proof of Claim:**
     - Suppose there exists \( i \geq 3 \) such that \( d_i \) is even.
     - Since \( d_i \) is a divisor of \( n \), \( \frac{d_i}{2} \) must also be a divisor of \( n \).
     - This implies \( \frac{d_i}{2} < d_1 \) or \( \frac{d_i}{2} > d_k \).
     - If \( \frac{d_i}{2} < d_1 \), then \( d_i < 2d_1 \). Since \( d_1 \) and \( d_2 \) are the smallest divisors, \( d_1 + d_2 = d_3 \). Thus, \( d_i < d_3 \), which contradicts \( i \geq 3 \).
     - If \( \frac{d_i}{2} > d_k \), then \( d_i > 2d_k \), which is impossible since \( d_i \leq d_k \).
     - Therefore, all \( d_i \) with \( i \geq 3 \) must be odd.
   - **Contradiction:**
     - For \( k \geq 5 \), consider \( d_5 = d_3 + d_4 \). Since \( d_3 \) and \( d_4 \) are both odd, their sum \( d_5 \) must be even.
     - This contradicts the claim that all \( d_i \) with \( i \geq 3 \) are odd.
   - Hence, no fibby numbers exist for \( k \geq 5 \).

The final answer is \( \boxed{ k = 3 } \) and \( k = 4 \).