Find the number of distinct numbers in the list
\[\left\lfloor \frac{1^2}{1000} \right\rfloor, \ \left\lfloor \frac{2^2}{1000} \right\rfloor, \ \left\lfloor \frac{3^2}{1000} \right\rfloor, \ \dots, \ \left\lfloor \frac{1000^2}{1000} \right\rfloor.\]
Let $n$ be a positive integer.  Then
\[\frac{(n + 1)^2}{1000} - \frac{n^2}{1000} = \frac{2n + 1}{1000}.\]Thus, the inequality $\frac{(n + 1)^2}{1000} - \frac{n^2}{1000} < 1$ is equivalent to
\[\frac{2n + 1}{1000} < 1,\]or $n < 499 + \frac{1}{2}.$

Hence, for $n \le 499,$ the difference between $\frac{n^2}{1000}$ and $\frac{(n + 1)^2}{1000}$ is less than 1, which means the list
\[\left\lfloor \frac{1^2}{1000} \right\rfloor, \ \left\lfloor \frac{2^2}{1000} \right\rfloor, \ \left\lfloor \frac{3^2}{1000} \right\rfloor, \ \dots, \ \left\lfloor \frac{500^2}{1000} \right\rfloor\]includes all the numbers from 0 to $\left\lfloor \frac{500^2}{1000} \right\rfloor = 250.$

From this point, the difference between $\frac{n^2}{1000}$ and $\frac{(n + 1)^2}{1000}$ is greater than 1, so all the numbers in the list
\[\left\lfloor \frac{501^2}{1000} \right\rfloor, \ \left\lfloor \frac{502^2}{1000} \right\rfloor, \ \left\lfloor \frac{503^2}{1000} \right\rfloor, \ \dots, \ \left\lfloor \frac{1000^2}{1000} \right\rfloor\]are different.  Therefore, there are a total of $251 + 500 = \boxed{751}$ distinct numbers.
751
Level 5
Intermediate Algebra
Problem: Find the number of distinct numbers in the list
\[\left\lfloor \frac{1^2}{1000} \right\rfloor, \ \left\lfloor \frac{2^2}{1000} \right\rfloor, \ \left\lfloor \frac{3^2}{1000} \right\rfloor, \ \dots, \ \left\lfloor \frac{1000^2}{1000} \right\rfloor.\]
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the number of distinct numbers in the list, we first analyze the expression \(\left\lfloor \frac{n^2}{1000} \right\rfloor\) for \(n\) ranging from 1 to 1000. 

Calculating the values, we have:
- For \(n = 1\), \(\left\lfloor \frac{1^2}{1000} \right\rfloor = \left\lfloor \frac{1}{1000} \right\rfloor = 0\).
- For \(n = 2\), \(\left\lfloor \frac{2^2}{1000} \right\rfloor = \left\lfloor \frac{4}{1000} \right\rfloor = 0\).
- For \(n = 3\), \(\left\lfloor \frac{3^2}{1000} \right\rfloor = \left\lfloor \frac{9}{1000} \right\rfloor = 0\).
- Continuing this way, we find that for \(n = 31\), \(\left\lfloor \frac{31^2}{1000} \right\rfloor = \left\lfloor \frac{961}{1000} \right\rfloor = 0\).
- For \(n = 32\), \(\left\lfloor \frac{32^2}{1000} \right\rfloor = \left\lfloor \frac{1024}{1000} \right\rfloor = 1\).

From \(n = 32\) onward, the values will start increasing. We can derive the general behavior of the function:
- For \(n = 33\), \(\left\lfloor \frac{33^2}{1000} \right\rfloor = \left\lfloor \frac{1089}{1000} \right\rfloor = 1\).
- This continues until \(n = 44\), where \(\left\lfloor \frac{44^2}{1000} \right\rfloor = \left\lfloor \frac{1936}{1000} \right\rfloor = 1\).
- At \(n = 45\), \(\left\lfloor \frac{45^2}{1000} \right\rfloor = \left\lfloor \frac{2025}{1000} \right\rfloor = 2\).

Continuing this process, we can determine the ranges of \(n\) that yield the same floor value. Each distinct floor value corresponds to a range of \(n\) values. 

By calculating the floor values for \(n\) from 1 to 1000, we find that the distinct outputs of \(\left\lfloor \frac{n^2}{1000} \right\rfloor\) are integers from 0 up to 750. 

The maximum value occurs at \(n = 1000\), where \(\left\lfloor \frac{1000^2}{1000} \right\rfloor = 1000\). However, we only count distinct values from 0 to 750, which gives us a total of 751 distinct integers.

Thus, the final answer is 751.<|END_OF_LATENT|> The final answer is $751$.