jmc

The decimal representation of a simplified fraction terminates if and only if the denominator is divisible by no primes other than 2 and 5. The prime factorization of $1375$ is $11\cdot 5^3$. For the fraction to simplify to having only the primes $2$ and $5$ in the denominator, there must be a factor of $11$ in the numerator. There are $\lfloor \frac{1000}{11}\rfloor =90$ multiples of $11$ between $1$ and $1000$, so there are $\boxed{90}$ integers values for $n$.