The answer of the problem is the set of all positive real numbers less than $2$. We consider two cases.

Case I We assume that $0 < a < 2$. Then there is a positive $n$ such that $2^{n-1} > a^n$. We define $A_n = \{m \mid m \text{ is a multiple of } 2^{n-1}\}$ and
$$A_i = \{2^{i-1}m \mid m \text{ is an odd integer}\},$$
for $1 \le i \le n-1$. Then $A_1, A_2, \dots, A_n$ is a partition of the set of positive integers satisfying the conditions of the problem.

Case II We assume that $a \ge 2$. We claim that no such partition exists. To prove by contradiction, we assume on the contrary that $A_1, A_2, \dots, A_n$ is a partition satisfying the conditions of the problem. Let $N = \{1, 2, \dots, 2^n\}$. For every $i$ with $1 \le i \le n$, let $B_i = A_i \cap N$. We assume that $B_i = \{b_1, b_2, \dots, b_m\}$ with $b_1 < b_2 < \dots < b_m$. We have
$$2^n > b_m - b_1 = (b_m - b_{m-1}) + (b_{m-1} - b_{m-2}) + \dots + (b_2 - b_1) \ge (m-1)2^i,$$
implying that $m-1 < 2^{n-i}$, or $m \le 2^{n-i}$. Since $A_1, A_2, \dots, A_n$ is a partition, $B_i \cap B_j = \emptyset$ for $1 \le i < j \le n$ and $N = B_1 \cup B_2 \cup \dots \cup B_n$. It follows that
$$2^n = |N| = |B_1| + |B_2| + \dots + |B_n| \le \sum_{i=1}^{n} 2^{n-i} = 2^n - 1,$$
which is impossible. Hence our assumption was wrong and such a partition does not exist for every positive integer $n$.