In the game of [i]Ring Mafia[/i], there are $2019$ counters arranged in a circle. $673$ of these counters are mafia, and the remaining $1346$ counters are town. Two players, Tony and Madeline, take turns with Tony going first. Tony does not know which counters are mafia but Madeline does. 

On Tony’s turn, he selects any subset of the counters (possibly the empty set) and removes all counters in that set. On Madeline’s turn, she selects a town counter which is adjacent to a mafia counter and removes it. Whenever counters are removed, the remaining counters are brought closer together without changing their order so that they still form a circle. The game ends when either all mafia counters have been removed, or all town counters have been removed.

Is there a strategy for Tony that guarantees, no matter where the mafia counters are placed and what Madeline does, that at least one town counter remains at the end of the game?

[i]Proposed by Andrew Gu[/i]
1. **Restate the problem and initial conditions:**
   - There are 2019 counters arranged in a circle.
   - 673 of these counters are mafia, and the remaining 1346 counters are town.
   - Tony and Madeline take turns, with Tony going first.
   - Tony can remove any subset of counters on his turn.
   - Madeline removes a town counter adjacent to a mafia counter on her turn.
   - The game ends when either all mafia counters or all town counters are removed.
   - The goal is to determine if Tony has a strategy that guarantees at least one town counter remains at the end of the game.

2. **Claim:**
   - If Madeline can win the game from a certain position, she can also win if some mafia counters are added and/or some town counters are deleted from the initial position.

3. **Proof of the Claim:**
   - If a mafia counter is added, Madeline can pretend it does not exist. If Tony selects it, she can ignore it.
   - Adding a mafia counter does not limit which counters she can take away since she can take away the counters on both sides of the mafia.
   - If a town counter is deleted, Madeline can pretend it still exists and play accordingly.
   - Deleting a town counter does not limit the counters she can take away for the same reason as above.

4. **Induction Hypothesis:**
   - Let \( P(n) \) be the assertion that Tony cannot deploy a successful strategy if there are \( 2n \) town counters and \( n \) mafia counters.

5. **Base Case:**
   - For \( P(1) \), we have 2 town counters and 1 mafia counter.
   - If Tony does not remove any counter, Madeline removes both town counters.
   - If Tony removes one counter, Madeline can arrange the counters so that Tony removes a town counter, and she removes the remaining town counter.
   - If Tony removes more than one counter, Madeline can arrange the counters so that he removes both town counters.
   - Thus, \( P(1) \) is true.

6. **Inductive Step:**
   - Assume \( P(k-1) \) is true. We wish to show \( P(k) \) is true.
   - If Tony's first move is to remove \( m \neq 0 \) counters, Madeline can arrange the counters such that all \( m \) counters are town counters.
   - If \( m = 1 \), Madeline removes a town counter and then the adjacent mafia counter.
   - If \( m > 1 \), Madeline can add back \( m-1 \) town counters and repeat the \( m = 1 \) case.
   - In both cases, the game reduces to \( P(k-1) \), which is true by induction.

7. **Case when Tony does not remove any counters:**
   - Madeline arranges the counters such that there exists a triplet \( T = (a, b, c) \) where \( b \) is a town counter, and \( \{a, c\} \) contains one town and one mafia counter.
   - Madeline removes \( b \). The counters \( a \) and \( c \) are now adjacent.
   - If Tony chooses any counters other than \( a \) or \( c \), Madeline rearranges the counters such that all selected counters are towns and adds back that many town counters.
   - If Tony did not choose either \( a \) or \( c \), Madeline removes both \( a \) and \( c \).
   - If Tony chose \( a \) but not \( c \), set \( a \) as the town and delete \( c \).
   - If Tony chose \( c \) but not \( a \), set \( c \) as the town and delete \( a \).
   - If Tony chose both, the problem reduces to \( P(k-1) \).

8. **Conclusion:**
   - By induction, Tony cannot guarantee that at least one town counter remains at the end of the game.

The final answer is \(\boxed{\text{no}}\).