To find the longest representation of the number $1992$ as $n = a_1 + a_2 + \cdots + a_k$ where $a_1 > a_2 > \cdots > a_k > 0$ and each $a_i$ is divisible by $a_{i+1}$, we can use the following steps:

1. Initial Setup:
Let $m_i = \frac{a_i}{a_{i+1}}$ and $a_{k+1} = 1$. This implies that each $a_i$ is a multiple of $a_{i+1}$.

2. Representation in Terms of Fractions:
We can express $n$ in terms of these fractions:
$$\frac{n}{a_1} = 1 + \frac{1}{m_1} + \frac{1}{m_1 m_2} + \cdots + \frac{1}{m_1 m_2 \cdots m_k}$$
Let $n_i = \sum_{j=i+1}^k a_j$. Then,
$$\frac{n_i}{a_i} = \frac{1}{m_i} + \frac{1}{m_i m_{i+1}} + \cdots + \frac{1}{m_i m_{i+1} \cdots m_k}$$

3. **Finding $m_i$**:
To find $m_i$, we need the smallest integer $m_i$ such that $m_i \geq \frac{a_i}{n_i}$. This can be written as:
$$m_i = \left\lceil \frac{a_i}{n_i} \right\rceil$$

4. Algorithm for Representation:
- Start with $a_1$ such that $\frac{n}{2} < a_1 \leq n$.
- Denote $n_1 = n - a_1$.
- Recursively define $m_i = \left\lceil \frac{a_i}{n_i} \right\rceil$, $a_{i+1} = \frac{a_i}{m_i}$, and $n_{i+1} = n_i - a_{i+1}$.
- Continue this process until $n_k = 0$.

5. Proof of Termination:
Let $\frac{n_i}{a_i} = r_i = \frac{P_i}{Q_i}$ where $(P_i, Q_i) = 1$. If $P_i = 1$, we take the last $m_i = Q_i$ and the process stops. Otherwise, $m_i = \left\lceil \frac{Q_i}{P_i} \right\rceil$, and:
$$\frac{P_i}{Q_i} - \frac{1}{m_i} = \frac{P_i m_i - Q_i}{m_i Q_i}$$
Clearly, $0 < P_{i+1} = \frac{P_i m_i - Q_i}{\gcd(m_i, Q_i)} < P_i$ and $Q_{i+1} = \frac{Q_i m_i}{\gcd(m_i, Q_i)}$. Therefore, the process will eventually stop.

6. **Application to $1992$**:
- Start with $a_1 = 1992$.
- $a_2 = \frac{1992}{2} = 996$.
- $a_3 = \frac{996}{2} = 498$.
- $a_4 = \frac{498}{2} = 249$.
- $a_5 = \frac{249}{3} = 83$.
- $a_6 = \frac{83}{83} = 1$.

Thus, the longest representation of $1992$ is:
$$1992 = 1992 + 996 + 498 + 249 + 83 + 1$$

The final answer is $\boxed{1992 = 1992 + 996 + 498 + 249 + 83 + 1}$.