Consider a regular $n$-gon with $n$ odd. Given two adjacent vertices $A_{1}$ and $A_{2},$ define the sequence $(A_{k})$ of vertices of the $n$-gon as follows: For $k\ge 3,\, A_{k}$ is the vertex lying on the perpendicular bisector of $A_{k-2}A_{k-1}.$ Find all $n$ for which each vertex of the $n$-gon occurs in this sequence.
1. **Define the function \( f \):**
   Let \( f \) be a function whose domain is \( \{A_1, A_2, \ldots, A_n\} \) such that:
   \[
   f(A_1) = 0, \quad f(A_2) = 1
   \]
   and for \( k \geq 3 \),
   \[
   f(A_k) = \frac{f(A_{k-1}) + f(A_{k-2})}{2}
   \]

2. **Inductive formula for \( f(A_k) \):**
   Using strong induction, we can show that:
   \[
   f(A_{k+2}) = \frac{2^{k+1} + (-1)^k}{3 \cdot 2^k}
   \]
   for \( k \geq -1 \).

3. **Condition for \( n \) being a power of 3:**
   Suppose \( n \) is a power of 3. Let \( p \neq 3 \) be a prime divisor of \( n \). Let \( m \) be an integer such that \( 3m \equiv_p 2 \). Suppose there is a vertex \( A_{k+2} \) such that \( f(A_{k+2}) \equiv_n m \).

4. **Contradiction for non-power of 3:**
   Therefore, we have:
   \[
   f(A_{k+2}) \equiv_p m \Leftrightarrow 2^{k+1} + (-1)^k \equiv_p 3m \cdot 2^k \Leftrightarrow (-1)^k \equiv_p 2^k (3m - 2) \Leftrightarrow (-1)^k \equiv_p 0
   \]
   which is impossible. Therefore, \( n \) must be a power of 3.

5. **Lemma:**
   \[
   \text{ord}_{3^k}(-2) = 3^{k-1}
   \]
   where \( k \) is a positive integer.

6. **Proof of Lemma:**
   For \( k = 1 \), it is trivial. Consider \( k \geq 2 \). Since \( 3^1 \parallel (-2) - 1 \) and \( 3^{k-1} \parallel 3^{k-1} \), then, lifting the exponent, we have \( 3^k \parallel (-2)^{3^{k-1}} - 1 \). Therefore, \(\text{ord}_{3^k}(-2) \mid 3^{k-1} \). Suppose \(\text{ord}_{3^k}(-2) \neq 3^{k-1} \). Then \(\text{ord}_{3^k}(-2) \mid 3^{k-2} \), therefore \( 3^k \mid (-2)^{3^{k-2}} \). Since:
   \[
   (-2)^{3^{k-1}} - 1 = ((-2)^{3^{k-2}} - 1)((-2)^{2 \cdot 3^{k-2}} + (-2)^{3^{k-2}} + 1)
   \]
   and \( 3^k \mid (-2)^{3^{k-2}} - 1 \) and \( 3 \mid (-2)^{2 \cdot 3^{k-2}} + (-2)^{3^{k-2}} + 1 \), then \( 3^{k+1} \mid (-2)^{3^{k-1}} - 1 \), a contradiction.

7. **Final proof:**
   Now let's prove that, for every integer \( m \), there is a vertex \( A_{k+2} \) such that \( f(A_{k+2}) \equiv_n m \). Therefore, we have:
   \[
   f(A_{k+2}) \equiv_n m \Leftrightarrow 2^{k+1} + (-1)^k \equiv_{3n} 3m \cdot 2^k \Leftrightarrow (-2)^k (3m - 2) \equiv_{3n} 1
   \]
   Note that \( 3m - 2 \) covers all the remainders modulo \( 3n \) that are congruent to 1 modulo 3. From the Lemma, \( (-2)^k \) covers all the remainders modulo \( 3n \) that are congruent to 1 modulo 3, from where the result follows.

Concluding, each vertex of the \( n \)-gon occurs in the sequence if and only if \( n \) is a power of 3.

The final answer is \( \boxed{ n } \) is a power of 3.