3
9
8
6
4

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$2$ and $5$ are coprime, so the numerator will not affect whether the expression is an integer. For example, no matter how many times a number is doubled, it will not change whether the product is divisible by $5$ . Therefore, we will only consider the denominator.

If $n$ is positive, the denominator will be a power of $5$ . If $n$ is negative, the denominator will be a power of $2$ . $4000 = 2^5 \cdot 5^3$ , so $n$ may not exceed these exponents. Therefore, $-5 \le n \le 3$ (including $0$ ), so there are $3-(-5)+1 = \boxed{9}$ possible values of $n$ .