Let $0<c<1$ and $n$ a positive integer. Alice and Bob are playing a game. Bob writes $n$ integers on the board, not all equal. On a player's turn, they erase two numbers from the board and write their arithmetic mean instead. Alice starts and performs at most $cn$ moves. After her, Bob makes moves until there are only two numbers left on the board. Alice wins if these two numbers are different, and otherwise, Bob wins.
For which values of $c$ does Alice win for all large enough $n$?
To determine for which values of \( c \) Alice wins for all large enough \( n \), we need to analyze the strategies of both players and the conditions under which Alice can ensure that the final two numbers are different.

1. **Alice's Strategy for \( c \geq \frac{1}{2} \)**:
   - Assume \( n \) is even for simplicity. Without loss of generality, we can shift and scale the integers so that not all integers are of the same parity.
   - Alice can merge two opposite-parity integers to get a half-integer. She continues merging these half-integers arbitrarily until she has made \( \frac{1}{2}n \) moves.
   - After \( \frac{1}{2}n \) moves, Bob is left with \( \frac{1}{2}n - 1 \) untouched integers and one number \( k \) whose \( \nu_2 \) (the 2-adic valuation) equals \( -\frac{1}{2}n \).
   - Bob cannot merge the remaining numbers to get a single number with \( \nu_2 \) at most \( -\frac{1}{2}n \) without involving \( k \). Merging with \( k \) always leaves a number with \( \nu_2 \) more than \( -\frac{1}{2}n \).
   - Therefore, when only two numbers remain, one will have \( \nu_2 \) at most \( -\frac{1}{2}n \) and the other will have \( \nu_2 \) more than this, ensuring they are different. Thus, Alice wins.

2. **Bob's Strategy for \( c = \frac{1}{2} - r \) for some \( r > 0 \)**:
   - Bob writes \( \lfloor \frac{1}{2}n \rfloor \) numbers equal to 0 and \( \lceil \frac{1}{2}n \rceil \) numbers equal to 1.
   - After Alice makes her moves, Bob considers each number \( m \) formed by merging \( t > \frac{2}{r} + 1 \) zeroes and ones. He forms \( 1 - m \) by swapping the roles of 0 and 1.
   - For numbers formed from merging at most \( \frac{2}{r} \) zeroes and ones, Bob merges numbers created with the same process and creates a single copy of \( 1 - m \) for each instance of \( m \).
   - Alice can make at most \( \frac{r}{4}n \) numbers from merging more than \( \frac{2}{r} + 1 \) zeroes and ones. Thus, at most \( (\frac{1}{2} - r)n + \frac{r}{4}n + C \) zeroes and ones are used, where \( C \) is a bounded number.
   - For sufficiently large \( n \), at least one 0 and one 1 remain unmerged. Bob can then merge each Alice-created \( m \) with his own \( 1 - m \), merge all remaining zeroes together, merge all remaining ones together, and finally merge the resulting 0 and 1 together.
   - This leaves every number equal to \( \frac{1}{2} \), ensuring Bob wins.

3. **Conclusion**:
   - For \( c \geq \frac{1}{2} \), Alice can ensure that the final two numbers are different.
   - For \( c < \frac{1}{2} \), Bob can ensure that the final two numbers are the same.

The final answer is \( \boxed{ c \geq \frac{1}{2} } \).