Let $n$ be an even positive integer. Alice and Bob play the following game. Before the start of the game, Alice chooses a set $S$ containing $m$ integers and announces it to Bob. The players then alternate turns, with Bob going first, choosing $i\in\{1,2,\dots, n\}$ that has not been chosen and setting the value of $v_i$ to either $0$ or $1$. At the end of the game, when all of $v_1,v_2,\dots,v_n$ have been set, the expression $$E=v_1\cdot 2^0  + v_2 \cdot 2^1 + \dots + v_n \cdot 2^{n-1}$$ is calculated. Determine the minimum $m$ such that Alice can always ensure that $E\in S$ regardless of how Bob plays.
1. **Understanding the Problem:**
   - Alice and Bob play a game where Bob sets the values of \( v_i \) to either 0 or 1.
   - The final value \( E \) is calculated as \( E = v_1 \cdot 2^0 + v_2 \cdot 2^1 + \dots + v_n \cdot 2^{n-1} \).
   - Alice wants to ensure that \( E \in S \) regardless of Bob's choices.
   - We need to determine the minimum size \( m \) of the set \( S \) such that Alice can always ensure \( E \in S \).

2. **Construction:**
   - Consider the set of all numbers where \( v_{2i} = v_{2i-1} \) for \( i = 1, 2, \ldots, \frac{n}{2} \).
   - This means that for each pair of bits, the bits are the same.
   - There are \( 2^{\frac{n}{2}} \) such numbers because each pair of bits can be either 00 or 11, and there are \( \frac{n}{2} \) pairs.

3. **Bound:**
   - Suppose \( m < 2^{\frac{n}{2}} \).
   - Bob can always choose an index \( i \) such that the value \( v_i \) appears less than \( 2^{\frac{n}{2}-1} \) times in the set \( S \).
   - This is because if \( m < 2^{\frac{n}{2}} \), then for any bit position, either 0 or 1 must appear less than \( 2^{\frac{n}{2}-1} \) times.
   - Bob can then choose the value that appears less frequently, ensuring that the resulting \( E \) is not in \( S \).

4. **Conclusion:**
   - Therefore, the minimum \( m \) such that Alice can always ensure \( E \in S \) is \( 2^{\frac{n}{2}} \).

The final answer is \( \boxed{ m = 2^{\frac{n}{2}} } \).