Solution:

The pairs satisfying the condition of the problem are those of the type $a=2^{h} \cdot (2x+1)$, $b=2^{k} \cdot (2y+1)$ where $h \neq k$, that is, the pairs such that the maximal power of $2$ dividing the two numbers is different.

First, we show that for pairs NOT of this type, that is, of the type $a=2^{h} \cdot (2x+1)$, $b=2^{h} \cdot (2y+1)$, there exists a coloring in which there are neither two numbers of color $A$ at distance $a$ nor two numbers of color $B$ at distance $b$. The coloring is the following:
$$\underbrace{A A A \ldots A A}_{2^{h}} \underbrace{B B B \ldots B B}_{2^{h}} \underbrace{A A A \ldots}_{2^{h}}$$
Indeed, in this sequence two numbers at distance $2^{h}$ always have different color, numbers at distance $2 \cdot 2^{h}$ always have the same color, numbers at distance $3 \cdot 2^{h}$ always have different color, and so on; in particular, therefore, numbers having distance an odd multiple of $2^{h}$ always have different color, which is what we wanted to show.

We now show that in the case $a=2^{h} \cdot (2x+1)$, $b=2^{k} \cdot (2y+1)$ with $h<k$ (without loss of generality), for every coloring there exist two numbers of color $A$ with difference $a$ or two numbers of color $B$ with difference $b$.

Suppose, for contradiction, that there exists a coloring for which this is not true, and consider a number $n$ having color $A$ (obviously there exists a number colored $A$, otherwise they would all be colored $B$ and there would exist two numbers colored $B$ at any distance); now, necessarily the number $n+a$ must be of color $B$, and for the same reason the numbers $n+a+b$ and $n+a-b$ must be of color $A$.

In general, moving (in either direction, provided we remain among the positive integers) from a number by a step of length $a+b$ or $a-b$ (the latter can also be negative, i.e., a step to the left), we end up at a number of the same color; starting from $n$, therefore, all positive numbers of the type $n+r(a+b)+s(a-b)$ with $r, s$ integers are of the same color $A$ as $n$. (Note that one can move first by taking forward steps and then backward steps, thus not using negative numbers.)

We now show that among these numbers there is also the number $n+a$, thus arriving at a contradiction. Let $d=\operatorname{MCD}(a+b, a-b)$. By Bézout's theorem, the numbers of the form $r(a+b)+s(a-b)$ are exactly the multiples of $d$, so it suffices to show that $d$ is a divisor of $a$. Since $d \mid a+b$ and $d \mid a-b$, it is evident that $d \mid (a+b)+(a-b)=2a$. It follows that $d=2^{\delta} d_{1}$, where $\delta \leq h+1$ and $d_{1} \mid 2x+1$. On the other hand, the maximal power of $2$ dividing both $a+b$ and $a-b$ is $2^{h}$, and therefore $\delta \leq h$, that is, $d \mid a$, as we wanted.