Let $q$ be a positive rational number. Two ants are initially at the same point $X$ in the plane. In the $n$-th minute $(n = 1,2,...)$ each of them chooses whether to walk due north, east, south or west and then walks the distance of $q^n$ metres. After a whole number of minutes, they are at the same point in the plane (not necessarily $X$), but have not taken exactly the same route within that time. Determine all possible values of $q$.

Proposed by Jeremy King, UK
1. **Initial Setup and Assumptions:**
   Let \( q \) be a positive rational number. Two ants start at the same point \( X \) in the plane. In the \( n \)-th minute (\( n = 1, 2, \ldots \)), each ant chooses to walk due north, east, south, or west, covering a distance of \( q^n \) meters. After a certain number of minutes, they meet at the same point in the plane but have not taken exactly the same route.

2. **Representation in the Complex Plane:**
   Consider the ants' movements in the complex plane. Let the \( i \)-th move of ant 1 be denoted by \( \varepsilon_i \) and the \( i \)-th move of ant 2 be denoted by \( \varepsilon'_i \). Clearly, \( \varepsilon_i, \varepsilon'_i \in \{\pm 1, \pm i\} \).

3. **Equating the Paths:**
   If the ants meet at the \( n \)-th minute, we have:
   \[
   \sum_{i=1}^n \varepsilon_i q^i = \sum_{i=1}^n \varepsilon_i' q^i
   \]
   Dividing both sides by \( q \), we get:
   \[
   \sum_{i=1}^n \varepsilon_i q^{i-1} = \sum_{i=1}^n \varepsilon_i' q^{i-1}
   \]
   Rearranging, we obtain:
   \[
   \sum_{i=1}^n (\varepsilon_i - \varepsilon_i') q^{i-1} = 0
   \]

4. **Conjugation and Coefficients:**
   Conjugating the equation, we note that for every \( \varepsilon_i \), \( 1 \) and \( -1 \) remain unchanged, while \( i \mapsto -i \) and \( -i \mapsto i \). Adding the conjugated equation to the original, we find that every coefficient of \( q^k \) is either \( 0, \pm 2 \), or \( \pm 4 \).

5. **Application of the Rational Root Theorem:**
   Since \( q \) is a positive rational number, by the Rational Root Theorem, the possible values of \( q \) are \( \{1, 2, 2^{-1}\} \).

6. **Verification of \( q = 1 \):**
   Clearly, \( q = 1 \) works because the ants can take different paths and still meet at the same point after any number of minutes.

7. **Verification of \( q = 2 \):**
   For \( q = 2 \), consider the left-right moves as \( i \) moves and the up-down moves as \( j \) moves. Since their paths are not exactly the same, at least one of the \( i \) or \( j \) moves must be different for the two ants. Suppose the \( i \) moves are different. We must then have:
   \[
   \pm 2^{x_1} \pm \cdots \pm 2^{x_r} = \pm 2^{y_1} \pm \cdots \pm 2^{y_s}
   \]
   for some \( r, s \) with \( \{x_i\} \ne \{y_i\} \) and \( x_1 < \cdots < x_r \) and \( y_1 < \cdots < y_s \). Taking the positive terms on one side and the negative on the other, we get a contradiction as two different binary numbers cannot be equal.

8. **Verification of \( q = 2^{-1} \):**
   For \( q = 2^{-1} \), we move in the reverse direction, i.e., start at the endpoint and move towards the starting point. This case reduces to the case \( q = 2 \), which we have shown is not possible.

Therefore, the only possible value for \( q \) is \( 1 \).

The final answer is \( \boxed{ q = 1 } \).