For real number $a$ denote by $[a]$ the greatest integer not exceeding $a$.
How many positive integers $n\leq 10000$ are there which is $[\sqrt{n}]$ is a divisor of $n$?

2012 Tokyo Institute of Techonolgy entrance exam, problem 2-B
1. Let \( n \) be a positive integer such that \( n \leq 10000 \).
2. Define \( k = \lfloor \sqrt{n} \rfloor \). This means \( k \) is the greatest integer not exceeding \( \sqrt{n} \), so \( k \leq \sqrt{n} < k+1 \).
3. Therefore, \( k^2 \leq n < (k+1)^2 \).
4. We need \( k \) to be a divisor of \( n \). This means \( n = mk \) for some integer \( m \).
5. Since \( k^2 \leq n < (k+1)^2 \), substituting \( n = mk \) gives \( k^2 \leq mk < (k+1)^2 \).
6. Dividing the entire inequality by \( k \), we get \( k \leq m < k + \frac{2k+1}{k} \).
7. Simplifying the right-hand side, we get \( k \leq m < k + 2 + \frac{1}{k} \).
8. Since \( m \) must be an integer, the possible values for \( m \) are \( k, k+1, k+2 \) (as \( \frac{1}{k} \) is less than 1 for all \( k \geq 1 \)).
9. Therefore, for each \( k \), there are three values of \( n \) that satisfy the condition: \( n = k^2, n = k^2 + k, n = k^2 + 2k \).
10. We need to count the number of such \( n \) for \( n \leq 10000 \).
11. The maximum value of \( k \) such that \( k^2 \leq 10000 \) is \( k = \lfloor \sqrt{10000} \rfloor = 100 \).
12. For each \( k \) from 1 to 100, there are 3 values of \( n \) that satisfy the condition.
13. Therefore, the total number of such \( n \) is \( 3 \times 100 = 300 \).

The final answer is \(\boxed{300}\).