1. Graph Representation and State Definition:
- Represent the problem as a graph $G$ where each sheep is a vertex and edges represent mutual friendships.
- Define the state of each vertex (sheep) as $1$ if the sheep is friends with the wolf and $0$ otherwise.

2. Wolf's Move and State Change:
- Each day, the wolf can remove one of its friends (a vertex with state $1$).
- When the wolf eats a sheep $A$:
- If a friend of $A$ is originally a friend of the wolf, it un-friends the wolf.
- If a friend of $A$ is originally not a friend of the wolf, it becomes a friend of the wolf.

3. Example with Star Graph:
- Consider a star graph with $n$ vertices, where one central vertex $v$ is connected to $n-1$ leaf vertices.
- If there are an odd number of vertices with state $1$, the wolf can remove all the leaf vertices with state $1$, and then remove $v$.
- Removing $v$ is possible because there were initially an odd number of vertices with state $1$.
- After removing $v$, the wolf can remove all remaining vertices since they all have state $1$.

4. Counting Initial States:
- The number of ways to choose initial states such that there are an odd number of vertices with state $1$ is $2^{n-1}$.

5. **Proof for $m \leq 2^{n-1}$:**
- Let $R$ denote the number of vertices with state $1$.
- We need to prove that if the wolf can remove all vertices, then initially the condition $R \equiv n + |E(G)| \pmod{2}$ holds.
- Consider the invariant $S = R + |V| + |E|$, where $|V|$ and $|E|$ are the number of vertices and edges left, respectively.
- After each move, the parity of $S$ does not change, and at the end, $S = 0$.
- Therefore, initially $S$ should be even, which implies $R \equiv n + |E(G)| \pmod{2}$.

6. Conclusion:
- The largest integer $m$ such that there exists an initial friendship structure where the wolf has $m$ different ways of choosing initial friends to ensure it can eat all the sheep is $2^{n-1}$.

The final answer is $\boxed{ 2^{n-1} }$.