Let $n$ be a positive integer. Alice and Bob play the following game. First, Alice picks $n + 1$ subsets $A_1,...,A_{n+1}$ of $\{1,... ,2^n\}$ each of size $2^{n-1}$. Second, Bob picks $n + 1$ arbitrary integers $a_1,...,a_{n+1}$. Finally, Alice picks an integer $t$. Bob wins if there exists an integer $1 \le i \le n + 1$ and $s \in A_i$ such that $s + a_i \equiv t$ (mod $2^n$). Otherwise, Alice wins.
Find all values of $n$ where Alice has a winning strategy.
1. **Translation of the Problem**:
   - Alice picks \( n+1 \) subsets \( A_1, A_2, \ldots, A_{n+1} \) of \( \{1, 2, \ldots, 2^n\} \), each of size \( 2^{n-1} \).
   - Bob picks \( n+1 \) integers \( a_1, a_2, \ldots, a_{n+1} \).
   - Bob wins if there exists an integer \( 1 \le i \le n+1 \) and \( s \in A_i \) such that \( s + a_i \equiv t \pmod{2^n} \) for some \( t \) chosen by Alice.

2. **Bob's Strategy**:
   - Bob's strategy is to cover as many residues as possible at each step.
   - He starts by picking \( a_1 = 0 \).

3. **Lemma**:
   - **Statement**: Let \( A \) and \( X \) be subsets of residues modulo \( 2^n \) with \( |A| = 2^{n-1} \). There exists an integer \( a \) such that \( A + a \) covers at least half of the elements of \( X \).
   - **Proof**:
     - Fix some \( x \in X \).
     - Pick a residue \( a \) modulo \( 2^n \) randomly.
     - Define the random variable \( \xi_x \) which is 1 if \( x \in A + a \) and 0 otherwise.
     - The probability that \( \xi_x = 1 \) is \( \frac{1}{2} \) because \( |A| = 2^{n-1} \).
     - Consider the random variable \( \xi := \sum_{x \in X} \xi_x \).
     - By the linearity of expectation:
       \[
       \mathbb{E}[\xi] = \sum_{x \in X} \mathbb{E}[\xi_x] = \sum_{x \in X} \frac{1}{2} = \frac{|X|}{2}.
       \]
     - This means that there exists some \( a \) such that \( \xi \geq \frac{|X|}{2} \), i.e., \( A + a \) covers at least half of the elements of \( X \). \(\blacksquare\)

4. **Bob's Steps**:
   - Bob chooses \( a_1 = 0 \) and let the set of uncovered residues be \( X_2 \) with \( |X_2| = 2^{n-1} \).
   - For each subsequent step \( i = 2, 3, \ldots, n+1 \), Bob chooses \( a_i \) such that \( A_i + a_i \) covers at least half of the elements of \( X_i \).
   - Let \( X_{i+1} \) be the set of still uncovered residues after step \( i \).

5. **Conclusion**:
   - After \( n+1 \) steps, Bob will have managed to cover all the residues modulo \( 2^n \).

The final answer is Bob always wins, so there is no value of \( \boxed{ n } \) for which Alice has a winning strategy.