, 2011

It is easy to exhibit such a partition: set
$$A_1 = \bigcup_{n=1}^{\infty} \{n(2n-1) + 1, n(2n-1) + 2, \dots, n(2n-1) + 2n\},$$
$$A_2 = \bigcup_{n=0}^{\infty} \{n(2n+1) + 1, n(2n+1) + 2, \dots, n(2n+1) + 2n+1\}.$$
Since each class has arbitrarily large gaps, it cannot contain an infinite arithmetic sequence.

To exhibit such a partition for the further question, we will rely on building a bijection that will allow us to "destroy" every single infinite arithmetic progression, within each of the two partition classes. Take any bijection
$$\phi: \{1, 2, \dots\} \rightarrow \{1, 2\} \times \{1, 2, \dots\} \times \{1, 2, \dots\}.$$
Define $A(a, r) = \{a + nr : n = 0, 1, \dots\}$ for $a, r \in \{1, 2, \dots\}$; these are all possible infinite arithmetic progressions made of positive integers. At step $s = 0$, both classes $A_1, A_2$ are empty. Denote by $v_{s+1}$ the next value to be distributed in the sets of the partition, so at step $s = 0$ take $v_1 = 1$. Now proceed in an algorithmic way.

At step $s \ge 1$, let be $\phi(s) = (c, a, r)$. There exists a least $k \ge 0$ such that $v_s + k \in A(a, r)$. Put $v_s, v_s + 1, \dots, v_s + k$ alternatively in $A_1, A_2$, starting with the one having the least maximum. If this ends with $v_s + k \in A_c$, then take $v_{s+1} = v_s + k + 1$; if not, then force $v_s + k \in A_c$ and put $v_s + k + 1$ in the other class, then take $v_{s+1} = v_s + k + 2$. Continue with the next step $s + 1$.

This ensures that neither of the arithmetic progressions $A(a, r)$ will be contained in any of the two classes $A_1, A_2$, while the differences between pairs of consecutive elements within each class is at most $3$.