For how many ordered pairs of positive integers $(a, b)$ such that $a \le 50$ is it true that $x^2 - ax + b$ has integer roots?
1. We start by noting that for the quadratic equation \(x^2 - ax + b\) to have integer roots, the roots must be integers \(i\) and \(j\) such that:
   \[
   x^2 - ax + b = (x - i)(x - j) = x^2 - (i + j)x + ij
   \]
   This implies that:
   \[
   a = i + j \quad \text{and} \quad b = ij
   \]

2. Given \(a \leq 50\), we need to find the number of ordered pairs \((a, b)\) such that \(i\) and \(j\) are positive integers and \(i \leq j\).

3. For a fixed \(a\), the pairs \((i, j)\) must satisfy \(i + j = a\). Since \(i \leq j\), we have:
   \[
   i \leq \frac{a}{2} \quad \text{and} \quad j = a - i
   \]
   Therefore, \(i\) can take any integer value from 1 to \(\left\lfloor \frac{a}{2} \right\rfloor\).

4. The number of such pairs \((i, j)\) for a fixed \(a\) is \(\left\lfloor \frac{a}{2} \right\rfloor\).

5. To find the total number of ordered pairs \((a, b)\) for \(a \leq 50\), we sum \(\left\lfloor \frac{a}{2} \right\rfloor\) from \(a = 1\) to \(a = 50\):
   \[
   \sum_{k=1}^{50} \left\lfloor \frac{k}{2} \right\rfloor
   \]

6. We calculate the sum:
   \[
   \sum_{k=1}^{50} \left\lfloor \frac{k}{2} \right\rfloor = 0 + 1 + 1 + 2 + 2 + \ldots + 24 + 24 + 25
   \]

7. Grouping the terms, we have:
   \[
   0 + 2 \cdot (1 + 2 + \ldots + 24) + 25
   \]

8. The sum of the first 24 positive integers is:
   \[
   1 + 2 + \ldots + 24 = \frac{24 \cdot 25}{2} = 300
   \]

9. Therefore, the total sum is:
   \[
   0 + 2 \cdot 300 + 25 = 0 + 600 + 25 = 625
   \]

The final answer is \(\boxed{625}\)