Find all pairs $ (x; y) $ of positive integers such that
$$xy | x^2 + 2y -1.$$
Given the problem, we need to find all pairs \((x, y)\) of positive integers such that 
\[ xy \mid x^2 + 2y - 1. \]

1. **Initial Divisibility Condition:**
   Since \(xy \mid x^2 + 2y - 1\), it implies that \(x \mid x^2 + 2y - 1\). Therefore, we can write:
   \[ x \mid x^2 + 2y - 1 \implies x \mid 2y - 1. \]
   This means there exists an integer \(k\) such that:
   \[ 2y - 1 = xk. \]
   Solving for \(y\), we get:
   \[ y = \frac{xk + 1}{2}. \]

2. **Substituting \(y\) into the Original Condition:**
   We substitute \(y = \frac{xk + 1}{2}\) into the condition \(xy \mid x^2 + 2y - 1\):
   \[ xy = x \cdot \frac{xk + 1}{2} = \frac{x^2 k + x}{2}. \]
   We need:
   \[ \frac{x^2 k + x}{2} \mid x^2 + 2 \left(\frac{xk + 1}{2}\right) - 1. \]
   Simplifying the right-hand side:
   \[ x^2 + 2 \left(\frac{xk + 1}{2}\right) - 1 = x^2 + xk + 1 - 1 = x^2 + xk. \]
   Therefore, we need:
   \[ \frac{x^2 k + x}{2} \mid x^2 + xk. \]

3. **Simplifying the Divisibility Condition:**
   Since \(\frac{x^2 k + x}{2} \mid x^2 + xk\), we can multiply both sides by 2 to clear the fraction:
   \[ x^2 k + x \mid 2(x^2 + xk). \]
   Simplifying the right-hand side:
   \[ 2(x^2 + xk) = 2x^2 + 2xk. \]
   Therefore, we need:
   \[ x^2 k + x \mid 2x^2 + 2xk. \]

4. **Analyzing the Divisibility:**
   We can rewrite the condition as:
   \[ x^2 k + x \mid 2x^2 + 2xk - (x^2 k + x) = 2x^2 + 2xk - x^2 k - x = 2x^2 + xk - x. \]
   Therefore, we need:
   \[ x^2 k + x \mid 2x^2 + xk - x. \]

5. **Considering Different Values of \(t\):**
   We have the equation:
   \[ x + k = yt. \]
   Substituting \(y = \frac{xk + 1}{2}\), we get:
   \[ x + k = t \left(\frac{xk + 1}{2}\right). \]
   Simplifying, we get:
   \[ 2(x + k) = t(xk + 1). \]
   \[ 2x + 2k = txk + t. \]

   - **Case \(t = 1\):**
     \[ 2x + 2k = xk + 1. \]
     \[ xk - 2x - 2k = -1. \]
     \[ (x - 2)(k - 2) = 3. \]
     The integer solutions are:
     \[ (x, k) = (3, 5), (5, 3). \]
     Substituting back, we get:
     \[ (x, y) = (3, 8), (5, 8). \]

   - **Case \(t = 2\):**
     \[ 2x + 2k = 2xk + 2. \]
     \[ xk - x - k = 1. \]
     \[ (x - 1)(k - 1) = 2. \]
     The integer solutions are:
     \[ (x, k) = (2, 2), (3, 1), (1, 3). \]
     Substituting back, we get:
     \[ (x, y) = (1, n), (2n - 1, n). \]

   - **Case \(t = 3\):**
     \[ 2x + 2k = 3xk + 3. \]
     \[ 3xk - 2x - 2k = 3. \]
     \[ \frac{2}{k} + \frac{2}{x} = 3 + \frac{3}{xk}. \]
     If \(x, k \geq 2\), then \(LHS < RHS\). Thus, at least one of them must be 1. After checking, no new solutions are found.

Hence, the solutions are:
\[ (x, y) = (3, 8), (5, 8), (1, n), (2n - 1, n). \]

The final answer is \( \boxed{ (3, 8), (5, 8), (1, n), (2n - 1, n) } \).