4. Find all positive real numbers $a$ such that there exists a positive integer $n$ and $n$ pairwise disjoint infinite sets $A_{1}, A_{2}, \cdots, A_{n}$ satisfying $A_{1} \cup A_{2} \cup \cdots \cup A_{n}=\mathbf{N}^{*}$, and for any two numbers $b>c$ in each $A_{i}$, we have $b-c \geqslant a^{i}$.
4. If $0a^{n}$. Let
$A_{i}=\left\{2^{i-1} m \mid m\right.$ 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 $\left|A_{i} \cap M\right| \leqslant 2^{n-i}$. Suppose $A_{i} \cap$ $M=\left\{x_{1}, x_{2}, \cdots, x_{m}\right\}, x_{1}x_{m}-x_{1}=\left(x_{m}-x_{m-1}\right)+\left(x_{m-1}-x_{m-2}\right)+\cdots+\left(x_{2}-x_{1}\right) \geqslant(m-1) 2^{i} \text {. }$
So $m-1<2^{n-i}$, 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 $2^{n}=|M|=\sum_{i=1}^{n}\left|A_{i} \cap M\right| \leqslant \sum_{i=1}^{n} 2^{n-i}=2^{n}-1$,
a contradiction.
Therefore, the required $a$ is all positive real numbers less than 2.