A sequence of natural numbers $c_1, c_2,\dots$ is called [i]perfect[/i] if every natural
number $m$ with $1\le m \le c_1 +\dots+ c_n$ can be represented as
$m =\frac{c_1}{a_1}+\frac{c_2}{a_2}+\dots+\frac{c_n}{a_n}$
Given $n$, find the maximum possible value of $c_n$ in a perfect sequence $(c_i)$.
1. **Define the sequence \( C_i \):**
   Let \( C_1 = 2 \) and \( C_i = 4 \times 3^{i-2} \) for \( i \geq 2 \). We claim that for each \( i \), \( C_i \) is the maximum possible value of \( c_i \).

2. **Base case for induction:**
   For \( i = 1 \), if \( c_1 > 1 \), then setting \( (m, n) = (c_1 - 1, 1) \) shows that \( a_1 = \frac{c_1}{c_1 - 1} \) is an integer. This happens exactly when \( c_1 = 2 \). Thus, \( c_1 \leq C_1 \).

3. **Inductive step:**
   Suppose that \( c_i \leq C_i \) for \( i = 1, 2, \ldots, k-1 \), where \( k \geq 2 \). We need to show that \( c_k \leq C_k \).

4. **Finding \( a_i \) for \( (m, n) = (c_k, k) \):**
   Clearly, \( a_n \geq 2 \). Then:
   \[
   c_n = \sum_{i=1}^n \frac{c_i}{a_i} \leq \frac{c_n}{2} + \sum_{i=1}^{n-1} C_i
   \]
   This implies:
   \[
   c_n \leq 2 \sum_{i=1}^{n-1} C_i = C_n
   \]
   This completes the inductive step and the proof of the claim that \( c_i \leq C_i \) for all \( i \).

5. **Proving the existence of \( c_i = C_i \):**
   We now prove by induction on \( n \) that we can have \( c_i = C_i \) for all \( i \) simultaneously.

6. **Base case for existence:**
   For \( n = 1 \), if \( m = 1 \), we can set \( a_1 = 2 \); if \( m = 2 \), we can set \( a_1 = 1 \).

7. **Inductive step for existence:**
   Assuming that the claim is true for \( n = 1, 2, \ldots, k-1 \), where \( k \geq 2 \), we prove it for \( n = k \).

8. **Handling different ranges of \( m \):**
   - If \( m = 1 \), we may set \( a_i = nC_i \) for \( i = 1, 2, \ldots, n \).
   - If \( 2 \leq m \leq \frac{C_n}{2} + 1 \), set \( a_n = C_n \).
   - If \( \frac{C_n}{2} + 1 \leq m \leq C_n \), set \( a_n = 2 \).
   - If \( C_n + 1 \leq m \leq \frac{3C_n}{2} = \sum_{i=1}^n C_i \), set \( a_n = 1 \).

9. **Ensuring the representation:**
   In each case, \( 1 \leq m - \frac{C_n}{a_n} \leq \frac{C_n}{2} \). Thus, by the induction hypothesis, we may choose positive integers \( a_1, a_2, \ldots, a_{n-1} \) such that:
   \[
   m - \frac{C_n}{a_n} = \sum_{i=1}^{n-1} \frac{C_i}{a_i}
   \]
   as desired. This completes the proof.

\(\blacksquare\)

The final answer is \( \boxed{ C_n = 4 \times 3^{n-2} } \) for \( n \geq 2 \) and \( C_1 = 2 \).