Let $r>1$ be a rational number. Alice plays the following game. Initially, on the real line, there is a red pawn at 0 and a blue one at 1. On each turn, Alice chooses an integer $k$, and notes $x, y$ as the positions of the two pawns in the order she wants. She then moves the pawn at $y$ to $y'$ such that $y'-x=r^k(y-x)$. Alice wins if she can move the red pawn to 1 in at most 2021 moves.

For which $r$ can Alice win?
Let $r = c / d$ with $c$ and $d$ coprime and $c > d$. Note already that the length formed by the segment red pawn-blue pawn is always a power of $r$ by immediate induction, and that the red pawn is always strictly to the left of the blue one.

Also note that if Alice chooses twice in a row for $x$ the abscissa of the same color point, and chooses $k$ and $k'$, Alice could just as well have chosen the same $x$, and $k + k'$ and obtained the same situation in one move.

Thus, if Alice can move the red pawn to 1 in at most 2021 moves, she can do so by alternately moving the red and blue pawns.

If she moves the red pawn first, up to changing the origin and the unit of the real line, we can assume that the red point is at 0 and the blue one at 1, and that on this line Alice can move the red pawn to 1 in at most 2020 moves. Thus, in all cases, if Alice can move the red pawn to the blue one in 1 in 2021 moves, she can do so in 2020 moves by alternating a blue and a red move.

Let $(a, b)$ be the abscissas of the red and blue pawns before the blue move and then the red move. If Alice chooses $k$, $a$ does not move during the first move but $b$ becomes $r^k(b - a) + a$. After the red move, if Alice chooses $l$, the abscissa of the red pawn becomes $r^l(a - (r^k(b - a) + a)) + r^k(b - a) + a = a + (r^l - r^{l+k})(b - a)$. Since $b - a$ is of the form $r^h$ for some integer $h$, $a$ is replaced by $a + r^{l+h} - r^{l+k+h}$.

Thus, after $k$ moves with a blue then a red, we obtain an abscissa of the form $\sum_{l=1}^{k} r^{a_l} - r^{b_l}$ for $(a_l)_{1 \leq l \leq k}$ and $(b_l)_{1 \leq l \leq k}$ integers. If Alice manages to place the red token at 1, then the previous equality holds for a good choice of $k \leq 1010$ (since she can do it in 200 moves) and of $(a_k)$ and $(b_k)$. Up to extending $(a_l)$ and $(b_l)$ by 0, the equation $\sum_{l=1}^{1010} r^{a_l} - r^{b_l} = 1$ has a solution which we now fix.

Ideally, we would like to use a bit of arithmetic, so consider modulo a number $n > 0$. The best would be that $r \equiv 1 \pmod{n}$, i.e., that $c \equiv d \pmod{n}$, which means that $n$ divides $c - d$. We can therefore look modulo $c - d$: note that $(c, c - d) = (d, c - d) = (d, c) = 1$, and $c - d > 0$. Thus $c$ and $d$ are coprime with $c - d$, so we can look at the equation modulo $c - d$, we get that $r \equiv \frac{d}{d} \equiv 1 \pmod{c - d}$, so $1 \equiv \sum_{l=1}^{1010} r^{a_l} - r^{b_l} \equiv \sum_{l=1}^{1010} 1 - 1 \equiv 0 \pmod{c - d}$. In particular, $c - d$ divides 1, so $c = d + 1$. Thus, $r = 1 + \frac{1}{d}$.

We can also look modulo $c + d$, for which $r \equiv -d / d \equiv -1 \pmod{c + d}$. In particular, $1 \equiv \sum_{l=1}^{1010} (-1)^{a_l} - (-1)^{b_l} \pmod{c + d}$. Since $\sum_{l=1}^{1010} (-1)^{a_l} - (-1)^{b_l}$ is between -2020 and 2020, and is 1 modulo $2d + 1$, we have $(2d + 1) + 1 \leq 2020$ or $-(2d + 1) + 1 \geq -2020$. In the first case, $2d \leq 2018$, so $d \leq 1009$ and in the second case, $2d \leq 2020$ so $d \leq 1010$. Thus $r$ is of the form $1 + \frac{1}{d}$ with $1 \leq d \leq 1010$.

Conversely, if $r$ is of this form, and we denote $(a, a+1)$ the positions of the red and blue pawns if they are at a distance of 2, by moving the blue pawn for $k=1$, its new position is $a + 1 + 1/d$. By moving the red pawn for $k=-1$, its new abscissa $a'$ satisfies: $a' - (a + 1 + 1/d) = \frac{(a - (a + 1 + 1/d))}{r} = -1$, so $a' = a + 1/d$ and each pawn has been translated by $1/d$.

Thus, by performing $k$ double moves, the two pawns end up at $(k/d, (k+1)/d)$. For $k = d$, $2d \leq 2021$ moves have been made, and the red pawn is at 1: the numbers of the form $1 + \frac{1}{d}$ with $1 \leq d \leq 1010$ work conversely.