Find all positive integer pairs $(a,b)$ for which the set of positive integers can be partitioned into sets $H_1$ and $H_2$ such that neither $a$ nor $b$ can be represented as the difference of two numbers in $H_i$ for $i=1,2$.
To solve the problem, we need to find all positive integer pairs \((a, b)\) such that the set of positive integers can be partitioned into sets \(H_1\) and \(H_2\) where neither \(a\) nor \(b\) can be represented as the difference of two numbers in \(H_i\) for \(i=1,2\).

1. **Power of 2 in Factorization**:
   Let \(P(n)\) denote the power of 2 in \(n\)'s factorization. We will prove that \(P(a) = P(b)\).

2. **Assume 1 is in \(H_1\)**:
   Suppose \(1 \in H_1\). Then, \(H_1\) must contain numbers of the form \(2ak + 1\) and \(2bn + 1\) for integers \(k\) and \(n\).

3. **Difference of Elements in \(H_1\)**:
   Consider the difference between two elements in \(H_1\):
   \[
   (2ak + 1) - (2bn + 1) = 2ak - 2bn = 2(ak - bn)
   \]
   For \(a\) to not be representable as the difference of two numbers in \(H_1\), \(2(ak - bn) \neq a\).

4. **Power Comparison**:
   Let's assume \(P(a) \geq P(b)\). Then, we have:
   \[
   2(ak - bn) = a \implies a = \frac{2bn}{2k - 1}
   \]
   For this equation to have no solutions, \(P(a)\) must equal \(P(b)\). If \(P(a) > P(b)\), there would be an infinite number of solutions.

5. **Conclusion**:
   Therefore, the only way to ensure that neither \(a\) nor \(b\) can be represented as the difference of two numbers in \(H_i\) is if \(P(a) = P(b)\).

The final answer is \( \boxed{ (a, b) } \) such that \(P(a) = P(b)\).