A pair of positive integers $m, n$ is called [i]guerrera[/i], if there exists positive integers $a, b, c, d$ such that $m=ab$, $n=cd$ and $a+b=c+d$. For example the pair $8, 9$ is [i]guerrera[/i] cause $8= 4 \cdot 2$, $9= 3 \cdot 3$ and $4+2=3+3$. We paint the positive integers if the following order:

We start painting the numbers $3$ and $5$. If a positive integer $x$ is not painted and a positive $y$ is painted such that the pair $x, y$ is [i]guerrera[/i], we paint $x$. 

Find all positive integers $x$ that can be painted.
To solve this problem, we need to determine which positive integers can be painted based on the given rules. We start by understanding the concept of a *guerrera* pair and then proceed to analyze the painting process.

1. **Definition of a *guerrera* pair**:
   A pair of positive integers \( (m, n) \) is called *guerrera* if there exist positive integers \( a, b, c, d \) such that:
   \[
   m = ab, \quad n = cd, \quad \text{and} \quad a + b = c + d
   \]
   For example, \( (8, 9) \) is a *guerrera* pair because \( 8 = 4 \cdot 2 \), \( 9 = 3 \cdot 3 \), and \( 4 + 2 = 3 + 3 \).

2. **Initial painted numbers**:
   We start by painting the numbers \( 3 \) and \( 5 \).

3. **Painting process**:
   If a positive integer \( x \) is not painted and there exists a painted positive integer \( y \) such that \( (x, y) \) is a *guerrera* pair, then we paint \( x \).

4. **Constructing the triangle**:
   We construct a triangle where the \( i \)-th row consists of integers \( n \) such that there exist \( a, b \in \mathbb{Z}^+ \) satisfying \( a + b = i + 1 \). This means the \( i \)-th row contains all products \( ab \) where \( a + b = i + 1 \).

5. **Painting the rows**:
   - The third row contains \( 3 \) (since \( 3 = 1 \cdot 3 \) and \( 1 + 3 = 4 \)).
   - The fourth row contains \( 4 \) (since \( 4 = 2 \cdot 2 \) and \( 2 + 2 = 4 \)).
   - The fifth row contains \( 5 \) (since \( 5 = 1 \cdot 5 \) and \( 1 + 5 = 6 \)).

6. **General case for \( i \geq 6 \)**:
   - If \( i \) is composite, then \( i = ab \) with \( a, b > 1 \). Since \( a + b \leq i \), there is an earlier row containing \( i \).
   - If \( i \) is prime, then \( i - 1 \) is not prime. The second entry in the row, \( 2(i-1) \), can be factored as \( ab \) with \( a, b > 2 \) whenever \( i \geq 7 \). Thus, \( i = 2 + (i - 2) \leq a + b \).

7. **Inductive argument**:
   Suppose all rows less than \( i \) are painted (excluding the first and second rows). Then some element of row \( i \) must be painted, and hence row \( i \) will also get painted.

8. **Conclusion**:
   By induction, all positive integers \( x \geq 3 \) will eventually be painted. The only integers that cannot be painted are \( 1 \) and \( 2 \).

The final answer is \(\boxed{\mathbb{Z}^+ \setminus \{1, 2\}}\).