A king decides to reward one of his knights by making a game. He sits the knights at a round table and has them call out $1,2,3,1,2,3,\dots$ around the circle (that is, clockwise, and each person says a number). The people who say $2$ or $3$ immediately lose, and this continues until the last knight is left, the winner.

Numbering the knights initially as $1,2,\dots,n$, find all values of $n$ such that knight $2008$ is the winner.
1. **Understanding the Problem:**
   - We have knights sitting in a circle, numbered from 1 to \( n \).
   - They call out numbers in the sequence \( 1, 2, 3, 1, 2, 3, \ldots \).
   - Knights who call out 2 or 3 are eliminated.
   - We need to find all values of \( n \) such that knight 2008 is the last one remaining.

2. **Initial Observations:**
   - The sequence of numbers called out is periodic with a period of 3.
   - Only knights who call out 1 survive each round.

3. **Claim 1:**
   - **Statement:** Knight \( K_1 \) is the winner if and only if \( n = 3^k \) or \( n = 2 \cdot 3^k \) for some non-negative integer \( k \).
   - **Proof by Induction:**
     - **Base Case:** For \( n \leq 3 \):
       - \( n = 1 \): \( K_1 \) wins.
       - \( n = 2 \): \( K_1 \) wins.
       - \( n = 3 \): \( K_1 \) does not win (all knights are eliminated).
     - **Inductive Step:**
       - Assume the claim holds for all \( n < N \) where \( N \geq 4 \).
       - Consider \( n = N \):
         - Knight \( K_n \) will say \( n \mod 3 \).
         - Knight \( K_1 \) will say \( (n+1) \mod 3 \).
         - \( K_1 \) says 1 again if and only if \( 3 \mid n \).
         - If \( 3 \mid n \), then \( n = 3m \) and there are \( \frac{n}{3} = m \) knights left.
         - By induction, \( K_1 \) wins if and only if \( m = 3^k \) or \( m = 2 \cdot 3^k \).
         - Therefore, \( n = 3^{k+1} \) or \( n = 2 \cdot 3^{k+1} \).

4. **Claim 2:**
   - **Statement:** Suppose \( m = 3t + 1 \) for some non-negative integer \( t \). Then \( K_m \) is the winner if and only if \( n = 2t + 3^k \) or \( n = 2t + 2 \cdot 3^k \) for some integer \( k \geq \lfloor \log_3(k) \rfloor \).
   - **Proof:**
     - When it's \( K_m \)'s turn to say 1, there are exactly \( n - 2k \) knights left.
     - \( K_m \) is now at the first position.
     - The claim follows from Claim 1.

5. **Application to Knight 2008:**
   - \( 2008 = 3 \times 669 + 1 \).
   - By Claim 2, the answer is \( n = 2 \times 669 + 3^k \) or \( n = 2 \times 669 + 2 \cdot 3^k \) for \( k \geq 6 \).

The final answer is \( \boxed{ n = 1338 + 3^k } \) or \( n = 1338 + 2 \cdot 3^k \) for \( k \geq 6 \).