Let $n \geq 3$ be a positive integer. In an election debate, we have $n$ seats arranged in a circle and these seats are numbered from $1$ to $n$, clockwise. In each of these chairs sits a politician, who can be a liar or an honest one. Lying politicians always tell lies, and honest politicians always tell the truth.

At one heated moment in the debate, they accused each other of being liars, with the politician in chair $1$ saying that the politician immediately to his left is a liar, the politician in chair $2$ saying that all the $2$ politicians immediately to his left are liars, the politician in the char $3$ saying that all the $3$ politicians immediately to his left are liars, and so on. Note that the politician in chair $n$ accuses all $n$ politicians (including himself) of being liars.

For what values of $n$ is this situation possible to happen?
1. **Define the problem and notation:**
   - Let \( n \geq 3 \) be a positive integer representing the number of seats arranged in a circle.
   - Each seat is numbered from \( 1 \) to \( n \) clockwise.
   - Each politician in a seat can either be a liar or an honest person.
   - Lying politicians always lie, and honest politicians always tell the truth.
   - Each politician accuses the politicians to their left of being liars, with the politician in seat \( n \) accusing all \( n \) politicians (including themselves).

2. **Initial observations:**
   - Denote \( T_r \) as the person in chair \( r \).
   - Let \( T_h \) be the highest value chair \( h \) an honest person sits in.

3. **Lemma: Either \( T_1 \) or \( T_2 \) must tell the truth.**
   - Suppose \( T_s \) is an honest person in chair \( s \).
   - If \( s \) is even, then \( T_{\frac{s}{2} - 1} \) tells the truth, and \( T_k \) for \( \frac{s}{2} - 1 < k < s \) are liars.
   - If \( s \) is odd, then \( T_{\lfloor \frac{s}{2} \rfloor} \) tells the truth, and \( T_k \) for \( \lfloor \frac{s}{2} \rfloor < k < s \) are liars.
   - Repeating this process, we eventually find that either \( T_1 \) or \( T_2 \) must be honest.

4. **Analyzing the range of \( h \):**
   - All \( T_k \) for \( h < k \leq n \) are liars.
   - This implies that some \( T_{k \pmod{n}} \) where \( h + 1 < k \leq 2h + 2 \pmod{n} \) must be a truth-teller, forcing some \( T_{k \pmod{n}} \) to be \( T_1 \) or \( T_2 \).
   - Therefore, \( 2h + 2 \geq n + 1 \) or \( h \geq \frac{n - 1}{2} \).

5. **Further constraints on \( h \):**
   - Since \( T_h \) is honest, all \( T_k \) where \( h < k \leq 2h \pmod{n} \) must be liars.
   - This means no \( T_k \) can be both \( T_1 \) and \( T_2 \), forcing \( 2h \leq n + 1 \) or \( h \leq \frac{n + 1}{2} \).

6. **Combining inequalities:**
   - Combining the inequalities, we get \( \frac{n - 1}{2} \leq h \leq \frac{n + 1}{2} \).
   - If \( n \) is even, \( h = \frac{n}{2} \).
   - If \( n \) is odd, \( h = \frac{n \pm 1}{2} \).

7. **Conclusion:**
   - If \( n \) is even, it is a possible value.
   - If \( n \) is odd, as long as \( h = \frac{n - 1}{2} \) leads to \( T_1 \) or \( h = \frac{n + 1}{2} \) leads to \( T_2 \) telling the truth, it will always be a possible value.
   - If we reverse the deduction process, we see that as long as \( n \neq a + (a + 1) = 2a + 1 \) where \( a = 2^p - 2 \) for some positive integer \( p \), it will always be a possible value.

The final answer is \( \boxed{ n \neq 2a + 1 } \) where \( a = 2^p - 2 \) for some positive integer \( p \).