To solve this problem, we need to find all pairs of positive integers $(x, y)$ such that if $a$ and $b$ are relatively prime positive divisors of $x^3 + y^3$, then $a + b - 1$ is also a divisor of $x^3 + y^3$.

Let's analyze the problem by considering the possible cases and simplifying the condition.

**Step 1: Simplification of $x^3 + y^3$**

The sum of cubes can be expressed as:
$$x^3 + y^3 = (x + y)(x^2 - xy + y^2).$$
For the given condition, we consider two relatively prime positive divisors $a$ and $b$ of $x^3 + y^3$. To satisfy the condition, $a + b - 1$ must also be a divisor of $x^3 + y^3$.

Step 2: Analyze Specific Cases

Consider special cases for $x$ and $y$:

1. **Case 1: $x = y$**
$$x^3 + y^3 = 2x^3.$$
The condition simplifies because the divisors of $2x^3$ are of the form $2d^3$, where $d$ is a divisor of $x$. Two relatively prime divisors can be taken as $1$ and $2x^3$ itself, making $1 + 2x^3 - 1 = 2x^3$ clearly a divisor. Therefore, pairs $(2^k, 2^k)$ for non-negative integers $k$ satisfy the condition.

2. **Case 2: $x = 2z, y = 3z$**
$$x^3 + y^3 = (2z)^3 + (3z)^3 = 8z^3 + 27z^3 = 35z^3.$$
Here, considering relative primality, the divisors can be selected as $5$ and $7$. It satisfies $5 + 7 - 1 = 11$ as a divisor of $35z^3$. Thus, pairs $(2 \cdot 3^k, 3^k)$ for non-negative integers $k$ are solutions.

3. **Case 3: $x = 3z, y = 2z$**
$$x^3 + y^3 = (3z)^3 + (2z)^3 = 27z^3 + 8z^3 = 35z^3.$$
Similarly, taking relatively prime divisors $5$ and $7$ satisfies the condition. Hence, pairs $(3^k, 2 \cdot 3^k)$ also satisfy the condition.

Conclusion:

The complete set of solutions for pairs $(x, y)$ in positive integers that satisfy the given condition are:
$$\boxed{(2^k, 2^k)}, \quad \boxed{(2 \cdot 3^k, 3^k)}, \quad \text{and} \quad \boxed{(3^k, 2 \cdot 3^k)}$$
for non-negative integers $k$.