There are three flies of negligible size that start at the same position on a circular track with circumference 1000 meters. They fly clockwise at speeds of 2, 6, and $k$ meters per second, respectively, where $k$ is some positive integer with $7\le k \le 2013$. Suppose that at some point in time, all three flies meet at a location different from their starting point. How many possible values of $k$ are there?

[i]Ray Li[/i]
1. We need to find the values of \( k \) such that the three flies meet at a location different from their starting point. The flies travel at speeds of 2, 6, and \( k \) meters per second, respectively, on a circular track with a circumference of 1000 meters.

2. Let \( t \) be the time in seconds when the flies meet again at a location different from their starting point. The positions of the flies at time \( t \) can be described by the following congruences:
   \[
   2t \equiv 6t \equiv kt \pmod{1000}
   \]

3. For the flies to meet at the same point, the positions must be congruent modulo 1000. This gives us the system of congruences:
   \[
   2t \equiv 6t \pmod{1000}
   \]
   \[
   6t \equiv kt \pmod{1000}
   \]

4. From the first congruence, we have:
   \[
   2t \equiv 6t \pmod{1000} \implies 4t \equiv 0 \pmod{1000} \implies t \equiv 0 \pmod{250}
   \]
   This implies that \( t \) must be a multiple of 250.

5. From the second congruence, we have:
   \[
   6t \equiv kt \pmod{1000} \implies t(k - 6) \equiv 0 \pmod{1000}
   \]
   Since \( t \equiv 0 \pmod{250} \), we can write \( t = 250m \) for some integer \( m \). Substituting this into the congruence, we get:
   \[
   250m(k - 6) \equiv 0 \pmod{1000}
   \]
   Dividing both sides by 250, we obtain:
   \[
   m(k - 6) \equiv 0 \pmod{4}
   \]

6. For this congruence to hold for any integer \( m \), \( k - 6 \) must be a multiple of 4. Therefore, we can write:
   \[
   k - 6 = 4n \implies k = 4n + 6
   \]
   for some integer \( n \).

7. Given the constraint \( 7 \le k \le 2013 \), we substitute \( k = 4n + 6 \) and solve for \( n \):
   \[
   7 \le 4n + 6 \le 2013
   \]
   Subtracting 6 from all parts of the inequality:
   \[
   1 \le 4n \le 2007
   \]
   Dividing by 4:
   \[
   \frac{1}{4} \le n \le 501.75
   \]
   Since \( n \) must be an integer, we have:
   \[
   1 \le n \le 501
   \]

8. Therefore, \( n \) can take any integer value from 1 to 501, inclusive. This gives us 501 possible values for \( n \), and hence 501 possible values for \( k \).

The final answer is \(\boxed{501}\)