Determine all pairs $\{a, b\}$ of positive integers with the property that, in whatever manner you color the positive integers with two colors $A$ and $B$, there always exist two positive integers of color $A$ having their difference equal to $a$ [b]or[/b] of color $B$ having their difference equal to $b$.
1. **Understanding the Problem:**
   We need to determine all pairs \(\{a, b\}\) of positive integers such that, no matter how we color the positive integers with two colors \(A\) and \(B\), there always exist two positive integers of color \(A\) having their difference equal to \(a\) or of color \(B\) having their difference equal to \(b\).

2. **Claim:**
   We claim that such a pair \((a, b)\) is good if and only if \(\nu_2(a) \neq \nu_2(b)\), where \(\nu_2(n)\) denotes the highest power of 2 dividing \(n\).

3. **Proof of the Claim:**
   - **If \(\nu_2(a) \neq \nu_2(b)\):**
     Assume without loss of generality that \(\nu_2(a) > \nu_2(b)\). Let \(\nu_2(a) = k\) and \(\nu_2(b) = m\) with \(k > m\). We can write \(a = a' \cdot 2^k\) and \(b = b' \cdot 2^m\) where \(a'\) and \(b'\) are odd integers.
     
     Consider the coloring of integers such that:
     - Integers of the form \(n + 2^m k\) (for \(k \in \mathbb{Z}\)) are colored with color \(A\) if \(k\) is even and with color \(B\) if \(k\) is odd.
     
     This ensures that any two integers of the same color differ by a multiple of \(2^m\). Since \(a\) is a multiple of \(2^k\) and \(k > m\), \(a\) cannot be the difference between two integers of the same color. Thus, the condition is satisfied.

   - **If \(\nu_2(a) = \nu_2(b)\):**
     Let \(\nu_2(a) = \nu_2(b) = k\). We can write \(a = a' \cdot 2^k\) and \(b = b' \cdot 2^k\) where \(a'\) and \(b'\) are odd integers.
     
     Consider the coloring of integers such that:
     - Integers \(1, 2, \ldots, 2^k\) are colored with color \(A\).
     - Integers \(n + 2^k\) are colored with color \(A\) if \(n\) is even and with color \(B\) if \(n\) is odd.
     
     This ensures that any two integers of the same color differ by a multiple of \(2^k\). Since both \(a\) and \(b\) are multiples of \(2^k\), there is no guarantee that the difference \(a\) or \(b\) will always be between two integers of the same color. Thus, the condition is not satisfied.

4. **Conclusion:**
   Therefore, the pairs \((a, b)\) that satisfy the given condition are those for which \(\nu_2(a) \neq \nu_2(b)\).

The final answer is \( \boxed{ \{a, b\} \text{ such that } \nu_2(a) \neq \nu_2(b) } \).