Problem 1489

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 $\boxed{501}$