We have nothing in common!

Notice that $\frac { a }{ b }$ is reducible with the given conditions if and only if $a$ is congruent to zero modulo $2$ or $5$ . (Because $1000={ 2 }^{ 3 }{ 5 }^{ 3 }$ )

Now, we want to find the number of fractions $\frac { a }{ 1000-a }$ such that $a$ is not divisible by two or five and $0<\frac { a }{ 1000-a } <1$ . This is just finding the number of integers $a$ from $1$ to $499$ , inclusive, so that $a$ is not divisible by two or five.

We subtract the number of multiples of two and five from the total, and add back the double counted numbers: $499-\left\lfloor \frac { 499 }{ 2 } \right\rfloor -\left\lfloor \frac { 499 }{ 5 } \right\rfloor +\left\lfloor \frac { 499 }{ 10 } \right\rfloor =200$ . So we have $200$ irreducible fractions that satisfy the conditions.

Another way to do this would be to notice that not being divisible by two or five is actually periodic in groups of $10$ , and go from there.

Clearly, $\dfrac ab = \dfrac a{1000-a}$ . Now the fraction is irreducible iff, $gcd\left(a, 1000-a\right) = gcd\left(a, 1000\right) = 1$ .
Since, $\phi\left(1000\right) = 400$ there are $400$ such $a$ that $\dfrac ab$ is irreducible. However, for only half of these $a < b$ .

Hence, the answer is $\boxed{200}$ .