Let $n$ be a positive integer. We call a $n$-tuple $(a_1, . . . , a_n)$ of positive integers [i]nice [/i] if
$\bullet$ $gcd (a_1, . . . , a_n) = 1$, and
$\bullet$ $a_i|a_{i-1} + a_{i+1}$, for all $i = 1, . . . , n$ (we define $a_0 = a_n$ and $a_{n+1} = a1$ here).
Find the maximal possible value of the sum $a_1 +...+ a_n$ if $(a_1, . . . , a_n)$ is a nice $n$-tuple.
1. **Define the Fibonacci Sequence:**
   Let $\{F_n\}$ denote the Fibonacci sequence with initial conditions $F_0 = 1$ and $F_1 = 1$, and the recurrence relation $F_n = F_{n-1} + F_{n-2}$ for $n \geq 2$.

2. **Identify the Problem Constraints:**
   We need to find the maximal possible value of the sum $a_1 + a_2 + \cdots + a_n$ for a nice $n$-tuple $(a_1, a_2, \ldots, a_n)$, where:
   - $\gcd(a_1, a_2, \ldots, a_n) = 1$
   - $a_i \mid (a_{i-1} + a_{i+1})$ for all $i = 1, 2, \ldots, n$ (with $a_0 = a_n$ and $a_{n+1} = a_1$).

3. **Consider the Maximal Element:**
   Suppose $a_i$ is a maximal element in the $n$-tuple such that $a_{i-1}$ is not maximal. Then, we must have $a_i = a_{i-1} + a_{i+1}$.

4. **Delete the Maximal Element:**
   If we delete $a_i$, the resulting $(n-1)$-tuple is still nice. This is because:
   - $a_{i+1} \equiv a_i \pmod{a_{i-1}}$ implies $a_{i-1} \mid (a_{i-2} + a_i) \implies a_{i-1} \mid (a_{i-2} + a_{i+1})$.
   - Similarly, $a_{i+1} \mid (a_{i-1} + a_{i+2})$.
   - The gcd condition is preserved: 
     \[
     1 = \gcd(a_1, \ldots, a_{i-1}, a_{i-1} + a_{i+1}, a_i, \ldots, a_n) = \gcd(a_1, \ldots, a_{i-1}, 0, a_{i+1}, \ldots, a_n) = \gcd(a_1, \ldots, a_{i-1}, a_{i+1}, \ldots, a_n).
     \]

5. **Base Case and Induction:**
   - For $n = 1$, the only nice 1-tuple is $(1)$.
   - For $n = 2$, the only nice 2-tuples are $(1, 2)$ and $(1, 1)$ up to cyclic permutations.
   - Assume the claim is true for $n$. For an $(n+1)$-tuple $T$, the maximal element is at most $F_{n+1}$, and the second-largest element is at most $F_n$.

6. **Inductive Step:**
   - If the claim holds for $n$, then for an $(n+1)$-tuple $T$, the maximal element is the sum of two elements from the $n$-tuple $T'$ obtained by deleting it. Hence, it is at most $F_{n+1}$.
   - The second-largest element in $T$ is still in $T'$, so it is at most $F_n$.

7. **Sum of Fibonacci Numbers:**
   By successively deleting maximal elements and applying the above claim, the maximum value of $a_1 + a_2 + \cdots + a_n$ is $F_1 + F_2 + \cdots + F_n$. By induction, it is easy to show that this equals $F_{n+2} - 2$.

The final answer is $\boxed{F_{n+2} - 2}$