Determine all positive real numbers $a$ such that there exists a positive integer $n$ and partition $A_1$, $A_2$, ..., $A_n$ of infinity sets of the set of the integers satisfying the following condition: for every set $A_i$, the positive difference of any pair of elements in $A_i$ is at least $a^i$.
1. **Understanding the Problem:**
   We need to determine all positive real numbers \( a \) such that there exists a positive integer \( n \) and a partition \( A_1, A_2, \ldots, A_n \) of infinite sets of the set of integers satisfying the condition: for every set \( A_i \), the positive difference of any pair of elements in \( A_i \) is at least \( a^i \).

2. **Upper Density Concept:**
   The upper density of a set \( K \) is defined as:
   \[
   \overline{d}(K) = \limsup_{m \to \infty} \frac{|K \cap \{1, 2, \ldots, m\}|}{m}
   \]
   It can be proven that for any sets \( K \) and \( L \):
   \[
   \overline{d}(K \cup L) \leq \overline{d}(K) + \overline{d}(L)
   \]

3. **Proof of Upper Density Inequality:**
   For any \( m \), we know that:
   \[
   |K \cap \{1, 2, \ldots, m\}| + |L \cap \{1, 2, \ldots, m\}|
   \]
   is the number of elements in \( \{1, 2, \ldots, m\} \) which are in \( K \) or \( L \), with overlaps counted twice. 
   \[
   |(K \cup L) \cap \{1, 2, \ldots, m\}|
   \]
   counts the same thing, but with overlaps counted only once. Therefore:
   \[
   |K \cap \{1, 2, \ldots, m\}| + |L \cap \{1, 2, \ldots, m\}| \geq |(K \cup L) \cap \{1, 2, \ldots, m\}|
   \]
   Dividing by \( m \) and taking the \( \limsup \):
   \[
   \frac{|K \cap \{1, 2, \ldots, m\}|}{m} + \frac{|L \cap \{1, 2, \ldots, m\}|}{m} \geq \frac{|(K \cup L) \cap \{1, 2, \ldots, m\}|}{m}
   \]
   Hence:
   \[
   \overline{d}(K \cup L) \leq \overline{d}(K) + \overline{d}(L)
   \]

4. **Working with Natural Numbers:**
   Assume \( A_i \subseteq \mathbb{N} \). In \( A_i \), the difference between any two consecutive terms is at least \( a^i \). Therefore, the size of \( A_i \) in \( \{1, 2, \ldots, m\} \) is at most:
   \[
   \left\lceil \frac{m}{a^i} \right\rceil
   \]
   Dividing by \( m \) and taking the \( \limsup \):
   \[
   \overline{d}(A_i) \leq \frac{1}{a^i}
   \]

5. **Summing Upper Densities:**
   Since \( \mathbb{N} = A_1 \cup A_2 \cup \ldots \cup A_n \):
   \[
   1 = \overline{d}(\mathbb{N}) = \overline{d}(A_1 \cup A_2 \cup \ldots \cup A_n) \leq \sum_{i=1}^n \overline{d}(A_i) \leq \sum_{i=1}^n \frac{1}{a^i}
   \]
   The series \( \sum_{i=1}^{\infty} \frac{1}{a^i} \) converges to:
   \[
   \frac{\frac{1}{a}}{1 - \frac{1}{a}} = \frac{1}{a-1}
   \]
   Therefore:
   \[
   1 \leq \sum_{i=1}^n \frac{1}{a^i} < \sum_{i=1}^{\infty} \frac{1}{a^i} = \frac{1}{a-1}
   \]
   This implies:
   \[
   a < 2
   \]

6. **Constructing the Sets \( A_i \):**
   We can prove that any \( a < 2 \) works. Let \( n \) be an integer such that \( 2^{n-1} > a^n \). For example, take:
   \[
   n = \left\lceil \frac{\log 2}{\log 2 - \log a} \right\rceil
   \]
   Define \( A_i \) as the set of all integers \( 2^{i-1} \mod 2^i \) for \( 1 \leq i \leq n-1 \), and let \( A_n \) be all numbers divisible by \( 2^{n-1} \). Clearly, any two consecutive numbers in \( A_i \) have a difference of \( 2^i > a^i \) for \( i \neq n \), and two consecutive numbers in \( A_n \) have a difference \( 2^{n-1} > a^n \) as required.

\(\blacksquare\)

The final answer is \( \boxed{ a < 2 } \)