Alice and Bob play the following game: Alice picks a set $A = \{1, 2, ..., n \}$ for some natural number $n \ge 2$. Then, starting from Bob, they alternatively choose one number from the set $A$, according to the following conditions: initially Bob chooses any number he wants, afterwards the number chosen at each step should be distinct from all the already chosen numbers and should differ by $1$ from an already chosen number. The game ends when all numbers from the set $A$ are chosen. Alice wins if the sum of all the numbers that she has chosen is composite. Otherwise Bob wins. Decide which player has a winning strategy.

Proposed by [i]Demetres Christofides, Cyprus[/i]
To determine which player has a winning strategy, we need to analyze the game for different values of \( n \) and identify patterns or strategies that guarantee a win for either Alice or Bob.

1. **Case \( n = 2 \)**:
   - Bob chooses \( 1 \). Alice must choose \( 2 \).
   - The sum of Alice's chosen numbers is \( 2 \), which is not composite.
   - Bob wins.

2. **Case \( n = 3 \)**:
   - Bob chooses \( 1 \). Alice must choose \( 2 \).
   - The sum of Alice's chosen numbers is \( 2 \), which is not composite.
   - Bob wins.

3. **Case \( n = 4 \)**:
   - Bob chooses \( 3 \). Alice can choose \( 2 \) or \( 4 \).
     - If Alice chooses \( 2 \), Bob chooses \( 1 \), and Alice must choose \( 4 \).
     - The sum of Alice's chosen numbers is \( 2 + 4 = 6 \), which is composite.
   - Alice wins.

4. **Case \( n = 5 \)**:
   - Bob chooses \( 2 \). Alice can choose \( 1 \) or \( 3 \).
     - If Alice chooses \( 1 \), Bob chooses \( 3 \), and Alice must choose \( 4 \).
     - The sum of Alice's chosen numbers is \( 1 + 4 = 5 \), which is not composite.
   - Bob wins.

5. **Case \( n = 6 \)**:
   - Bob chooses \( 2 \). Alice can choose \( 1 \) or \( 3 \).
     - If Alice chooses \( 1 \), Bob chooses \( 3 \), and Alice must choose \( 4 \).
     - The sum of Alice's chosen numbers is \( 1 + 4 = 5 \), which is not composite.
   - Bob wins.

6. **Case \( n = 7 \)**:
   - Bob chooses \( 2 \). Alice can choose \( 1 \) or \( 3 \).
     - If Alice chooses \( 1 \), Bob chooses \( 3 \), and Alice must choose \( 4 \).
     - The sum of Alice's chosen numbers is \( 1 + 4 = 5 \), which is not composite.
   - Bob wins.

7. **Case \( n = 8 \)**:
   - Bob chooses \( 2 \). Alice can choose \( 1 \) or \( 3 \).
     - If Alice chooses \( 1 \), Bob chooses \( 3 \), and Alice must choose \( 4 \).
     - The sum of Alice's chosen numbers is \( 1 + 4 = 5 \), which is not composite.
   - Bob wins.

From the above analysis, we can see that Bob has a winning strategy for \( n = 2, 3, 5, 6, 7 \). However, for \( n = 4 \), Alice has a winning strategy. 

To generalize, we need to consider the parity of \( n \) and the initial choice of Bob. If Bob can force Alice into a position where her sum is not composite, he wins. Otherwise, Alice wins.

The final answer is Bob has a winning strategy for \( \boxed{ n = 2, 3, 5, 6, 7 } \), and Alice has a winning strategy for \( n = 4 \).