Berkeley Math Circle

Solution:
The answer is yes.
We will prove the following statement by induction: It is possible to color the rational numbers with denominator at most $n$ red and blue such that the conditions (i)-(iii) hold whenever the two rational numbers in question both have denominator at most $n$, and such that no number reverses color at any step. (The last condition ensures that, when the induction is complete, every number will have a well-defined color.)

We make two base cases. For $n=1$, color all the positive integers red and all the nonpositive integers blue. Then, for $n=2$, color the number $x+\frac{1}{2}$ (for integer $x$) blue if $x \geq 0$ and red if $x<0$. It is easy to see that conditions (i)-(iii) are satisfied so far.

Assume that all rational numbers with denominator less than $n$ ($n \geq 3$) have been colored. We must color the fractions $m / n$, where $m$ is an integer coprime to $n$. First, when $0<m<n$, color $m / n$ the opposite of $n / m$ (which has already been colored). Then color $1+m / n$ the same color as $m / n$ for each $m>0$ coprime to $n$. Finally, color $-m / n$ the opposite of $m / n$. Condition (i) is obviously satisfied. Condition (iii), that $m / n$ and $n / m$ are opposite colors, is relevant only if $|m|<n$; it is satisfied by definition if $m>0$, and otherwise it is enough to note that the row of numbers
$$\frac{m}{n}, \quad \frac{n}{m}, \quad-\frac{n}{m}, \quad-\frac{m}{n}$$
has been colored alternately red and blue, implying that the two end terms have opposite colors. It remains to verify condition (ii). If $x>1$ or $x<0$, this follows from (i) and the fact that $y$ and $y+1$ have been colored alike for $y>0$. So we are left with the case $0<x<1$, that is, proving that $m / n$ and $(n-m) / n$ are the same color when $0<m<n$. The six numbers
$$\frac{m}{n}, \frac{n}{m},-\frac{n-m}{m},-\frac{m}{n-m}, \frac{n}{n-m}, \frac{n-m}{n}$$
have been colored alternately red and blue by (ii) and (iii), so the two end terms have opposite colors.