For how many ordered pairs of positive integers $(a,b)$ with $a,b<1000$ is it true that $a$ times $b$ is equal to $b^2$ divided by $a$? For example, $3$ times $9$ is equal to $9^2$ divided by $3$.

[i]Ray Li[/i]
1. We start with the given equation:
   \[
   ab = \frac{b^2}{a}
   \]
2. Multiply both sides by \(a\) to eliminate the fraction:
   \[
   a^2 b = b^2
   \]
3. Assuming \(b \neq 0\), we can divide both sides by \(b\):
   \[
   a^2 = b
   \]
4. We need to find the number of ordered pairs \((a, b)\) such that \(a\) and \(b\) are positive integers and \(a, b < 1000\). From \(a^2 = b\), \(b\) must be a perfect square.
5. Since \(b < 1000\), we need \(a^2 < 1000\). The largest integer \(a\) such that \(a^2 < 1000\) is:
   \[
   a = \lfloor \sqrt{1000} \rfloor = \lfloor 31.62 \rfloor = 31
   \]
6. Therefore, \(a\) can take any integer value from 1 to 31. For each \(a\), \(b = a^2\) is a unique perfect square less than 1000.

Conclusion:
The number of ordered pairs \((a, b)\) is equal to the number of possible values for \(a\), which is 31.

The final answer is \(\boxed{31}\).