The Bank of Zürich issues coins with an $H$ on one side and a $T$ on the other side. Alice has $n$ of these coins arranged in a line from left to right. She repeatedly performs the following operation: if some coin is showing its $H$ side, Alice chooses a group of consecutive coins (this group must contain at least one coin) and flips all of them; otherwise, all coins show $T$ and Alice stops. For instance, if $n = 3$, Alice may perform the following operations: $THT \to HTH \to HHH \to TTH \to TTT$. She might also choose to perform the operation $THT \to TTT$.

For each initial configuration $C$, let $m(C)$ be the minimal number of operations that Alice must perform. For example, $m(THT) = 1$ and $m(TTT) = 0$. For every integer $n \geq 1$, determine the largest value of $m(C)$ over all $2^n$ possible initial configurations $C$.

[i]Massimiliano Foschi, Italy[/i]
1. **Define the Problem and Key Terms:**
   - We have \( n \) coins, each showing either \( H \) (heads) or \( T \) (tails).
   - Alice can flip any group of consecutive coins.
   - The goal is to determine the maximum number of operations \( m(C) \) required to turn any initial configuration \( C \) into all tails.

2. **Define a Run:**
   - A **run** is a contiguous sequence of coins that are either all heads or all tails.
   - The key observation is that each operation can eliminate one run of heads by flipping it to tails.

3. **Claim:**
   - The maximum number of runs of heads in any configuration of \( n \) coins is \( \lceil \frac{n}{2} \rceil \).

4. **Proof of the Claim:**
   - If there were more than \( \lceil \frac{n}{2} \rceil \) runs of heads, the sequence of coins would need to switch from heads to tails (or vice versa) at least \( 2 \lceil \frac{n}{2} \rceil \) times.
   - Since \( 2 \lceil \frac{n}{2} \rceil \geq n \), this is a contradiction because the total number of coins is \( n \).
   - Therefore, the maximum number of runs of heads is \( \lceil \frac{n}{2} \rceil \).

5. **Upper Bound on Operations:**
   - Each operation can eliminate one run of heads.
   - Thus, the maximum number of operations required is \( \lceil \frac{n}{2} \rceil \).

6. **Constructing a Configuration to Achieve the Maximum:**
   - For odd \( n \), consider the sequence \( C = HTHT\cdots HTH \).
     - Extend \( C \) to \( C' = THTHT\cdots HTHT \) by appending tails to both ends.
     - This sequence has \( \frac{n+1}{2} \) runs of heads, requiring \( \lceil \frac{n}{2} \rceil \) operations.
   - For even \( n \), consider the sequence \( C = HTHT\cdots HT \).
     - Extend \( C \) to \( C' = THTHT\cdots HTT \) by appending a tail to one end.
     - This sequence has \( \frac{n}{2} \) runs of heads, requiring \( \lceil \frac{n}{2} \rceil \) operations.

7. **Conclusion:**
   - The maximum number of operations \( m(C) \) required for any configuration \( C \) of \( n \) coins is \( \lceil \frac{n}{2} \rceil \).

The final answer is \( \boxed{\lceil \frac{n}{2} \rceil} \)