NMO Selection Tests for the Junior Balkan Mathematical Olympiad

The answer is in the affirmative. For example, one can consider
$$A = \{1, 3, 4, 7, 8, 9, 13, 14, 15, 16, 21, \dots\},$$
$$B = \{2, 5, 6, 10, 11, 12, 17, 18, 19, 20, 26, \dots\},$$
i.e., for the infinite word $W = w_1w_2\cdots w_k \cdots = aba^2b^2\cdots a^nb^n\cdots$, put $A$ the set of the indices $i$ for which $w_i = a$, and $B$ the set of the indices $i$ for which $w_i = b$. The terms of an arithmetic progression are equally spaced, while $A$ and $B$ both contain arbitrarily long gaps, hence the conclusion follows readily.