Solution:

For a given positive integer $n$, consider an arrangement of $n$ tokens in the plane, where the tokens are at points $A_{1}, A_{2}, \ldots, A_{n}$. Let $G$ be the centroid of the $n$ points, so as vectors (after an arbitrary choice of origin),
$$\vec{G}=\frac{\vec{A}_{1}+\vec{A}_{2}+\cdots+\vec{A}_{n}}{n}$$
Note that any move leaves the centroid $G$ unchanged. Therefore, if all the tokens are eventually moved to the same point, then this point must be $G$.

First we prove that if $n=2^{k}$ for some nonnegative integer $k$, then all $n$ tokens can always be eventually moved to the same point. We shall use induction on $k$.

The result clearly holds for $n=2^{0}=1$. Assume that it holds when $n=2^{k}$ for some nonnegative integer $k$. Consider a set of $2^{k+1}$ tokens at $A_{1}, A_{2}, \ldots, A_{2^{k+1}}$. Let $M_{i}$ be the midpoint of $A_{2 i-1}$ and $A_{2 i}$ for $1 \leq i \leq 2^{k}$.

First we move the tokens at $A_{2 i-1}$ and $A_{2 i}$ to $M_{i}$, for $1 \leq i \leq 2^{k}$. Then, there are two tokens at $M_{i}$ for all $1 \leq i \leq 2^{k}$. If we take one token from each of $M_{1}, M_{2}, \ldots, M_{2^{k}}$, then by the induction hypothesis, we can move all of them to the same point, say $G$. We can do the same with the remaining tokens at $M_{1}, M_{2}, \ldots, M_{2^{k}}$. Thus, all $2^{k+1}$ tokens are now at $G$, which completes the induction argument.

(Here is an alternate approach to the induction step: Given the tokens at $A_{1}, A_{2}, \ldots, A_{2^{k+1}}$, move the first $2^{k}$ tokens to one point $G_{1}$, and move the remaining $2^{k}$ tokens to one point $G_{2}$. Then $2^{k}$ more moves can bring all the tokens to the midpoint of $G_{1}$ and $G_{2}$.)

Now, assume that $n$ is not a power of 2. Take any line in the plane, and number it as a real number line. (Henceforth, when we refer to a token at a real number, we mean with respect to this real number line.)

At the start, place $n-1$ tokens at $0$ and one token at $1$. We observed that if we can move all the tokens to the same point, then it must be the centroid of the $n$ points. Here, the centroid is at $\frac{1}{n}$. We now prove a lemma.

Lemma. The average of any two dyadic rationals is also a dyadic rational. (A dyadic rational is a rational number that can be expressed in the form $\frac{m}{2^{a}}$, where $m$ is an integer and $a$ is a nonnegative integer.)

Proof. Consider two dyadic rationals $\frac{m_{1}}{2^{a_{1}}}$ and $\frac{m_{2}}{2^{a_{2}}}$. Then their average is
$$\frac{1}{2}\left(\frac{m_{1}}{2^{a_{1}}}+\frac{m_{2}}{2^{a_{2}}}\right)=\frac{1}{2}\left(\frac{2^{a_{2}} \cdot m_{1}+2^{a_{1}} \cdot m_{2}}{2^{a_{1}} \cdot 2^{a_{2}}}\right)=\frac{2^{a_{2}} \cdot m_{1}+2^{a_{1}} \cdot m_{2}}{2^{a_{1}+a_{2}+1}}$$
which is another dyadic rational.

On this real number line, a move corresponds to taking a token at $x$ and a token at $y$ and moving both of them to $\frac{x+y}{2}$, the average of $x$ and $y$. At the start, every token is at a dyadic rational (namely $0$ or $1$), which means that after any number of moves, every token must still be at a dyadic rational.

But $n$ is not a power of 2, so $\frac{1}{n}$ is not a dyadic rational. (Indeed, if we could express $\frac{1}{n}$ in dyadic form $\frac{m}{2^{a}}$, then we would have $2^{a}=m n$, which is impossible unless $m$ and $n$ are powers of 2.) This means that it is not possible for any token to end up at $\frac{1}{n}$, let alone all $n$ tokens.

We conclude that we can always move all $n$ tokens to the same point if and only if $n$ is a power of 2.