Starting with any $n$-tuple $R_0$, $n\ge 1$, of symbols from $A,B,C$, we define a sequence $R_0, R_1, R_2,\ldots,$ according to the following rule: If $R_j= (x_1,x_2,\ldots,x_n)$, then $R_{j+1}= (y_1,y_2,\ldots,y_n)$, where $y_i=x_i$ if $x_i=x_{i+1}$ (taking $x_{n+1}=x_1$) and $y_i$ is the symbol other than $x_i, x_{i+1}$ if $x_i\neq x_{i+1}$. Find all positive integers $n>1$ for which there exists some integer $m>0$ such that $R_m=R_0$.
1. **Counterexample for even \( n \):**
   Let's consider \( n = 2 \) and the initial tuple \( R_0 = (A, B) \).
   - For \( R_0 = (A, B) \):
     - \( x_1 = A \)
     - \( x_2 = B \)
     - \( x_3 = x_1 = A \)
   - According to the rule:
     - \( y_1 = \) symbol other than \( A \) and \( B \) (which is \( C \))
     - \( y_2 = \) symbol other than \( B \) and \( A \) (which is \( C \))
   - Thus, \( R_1 = (C, C) \).

   Now, let's find \( R_2 \):
   - For \( R_1 = (C, C) \):
     - \( x_1 = C \)
     - \( x_2 = C \)
     - \( x_3 = x_1 = C \)
   - According to the rule:
     - \( y_1 = C \) (since \( x_1 = x_2 \))
     - \( y_2 = C \) (since \( x_2 = x_3 \))
   - Thus, \( R_2 = (C, C) \).

   We see that \( R_1 = R_2 \), and it will remain \( (C, C) \) for all subsequent steps. Therefore, \( R_m \neq R_0 \) for any \( m > 0 \). This shows that for \( n = 2 \), there is no \( m > 0 \) such that \( R_m = R_0 \).

   By similar reasoning, for any even \( n \), we can construct a counterexample where the sequence does not return to the initial tuple.

2. **Injectivity for odd \( n \):**
   Let's consider an odd \( n \)-tuple \( R = (x_1, x_2, \ldots, x_n) \). We need to show that there is a unique \( n \)-tuple \( Q \) such that under the given operation, \( Q \) transforms to \( R \).

   Suppose \( R = (x_1, x_2, \ldots, x_n) \) and \( Q = (y_1, y_2, \ldots, y_n) \) such that \( R \) is obtained from \( Q \) by the given rule. We need to determine \( Q \) uniquely from \( R \).

   - If \( x_i = x_{i+1} \), then \( y_i = x_i \).
   - If \( x_i \neq x_{i+1} \), then \( y_i \) is the symbol other than \( x_i \) and \( x_{i+1} \).

   Since \( n \) is odd, the sequence of transformations will eventually cycle back to the original tuple due to the injective nature of the transformation. This means that for any odd \( n \), there exists a unique \( n \)-tuple \( Q \) that transforms to \( R \).

3. **Conclusion using Pigeonhole Principle:**
   Since the transformation is injective for odd \( n \), and there are only a finite number of possible \( n \)-tuples (specifically \( 3^n \) possible tuples), by the Pigeonhole Principle, the sequence must eventually repeat. Therefore, for odd \( n \), there exists some integer \( m > 0 \) such that \( R_m = R_0 \).

The final answer is \( \boxed{ n } \) is odd.