4. Find all positive real numbers $a$ such that there exists a positive integer $n$ and $n$ pairwise disjoint infinite sets of integers $A_{1}$, $A_{2}$, \cdots, $A_{n}$ satisfying $A_{1} \cup A_{2} \cup \cdots \cup A_{n}=\mathbf{Z}$, and for any two numbers $b>c$ in each $A_{i}$, we have
$$
b-c \geqslant a^{i} .
$$
(Yuan Hanhui)
4. If $0 < a < 2$, let $A_{i}=\left\{2^{i-1} \mathrm{~m} \mid \mathrm{m}\right.$ is odd $\}, i=1,2, \cdots, n-1$, $A_{n}=\left\{2^{n-1}\right.$ multiples $\}$.
Then this partition satisfies the requirements.
If $a \geqslant 2$, suppose $A_{1}, A_{2}, \cdots, A_{n}$ satisfy the requirements.
Let $M=\left\{1,2, \cdots, 2^{n}\right\}$.
We will prove $\left|A_{i} \cap M\right| \leqslant 2^{n-i}$.
Assume $A_{i} \cap M=\left\{x_{1}, x_{2}, \cdots, x_{m}\right\}, x_{1} < x_{2} < \cdots < x_{m}$, then
$$
\begin{aligned}
x_{m}-x_{1} & =\left(x_{m}-x_{m-1}\right)+\left(x_{m-1}-x_{m-2}\right)+\cdots+\left(x_{2}-x_{1}\right) \\
& \geqslant(m-1) 2^{i} .
\end{aligned}
$$

Therefore, $m-1<2^{n-i}$, i.e., $m<2^{n-i}+1$.
Hence $m \leqslant 2^{n-i}$.
$A_{i} \cap M(i=1,2, \cdots, n)$ is a partition of $M$, so
$$
2^{n}=|M|=\sum_{i=1}^{n}\left|A_{i} \cap M\right| \leqslant \sum_{i=1}^{n} 2^{n-i}=2^{n}-1 \text {, }
$$

which is a contradiction.
Therefore, the required $a$ is all positive real numbers less than 2.