Answer: 128

Small Hint:

If $\frac{a}{b}$ is in lowest terms and $ab=20!$, each full prime power of $20!$ must go entirely to one of $a$ or $b$

Big Hint:

Count ordered allocations of the distinct prime-power factors, then use the condition $a<b$

Solution:

The distinct primes dividing $20!$ are $2,$ $3,$ $5,$ $7,$ $11,$ $13,$ $17,$ and $19.$ If $\frac{a}{b}$ is in lowest terms and $ab=20!,$ the entire power of each of these eight primes must be assigned to either $a$ or $b.$ Thus there are $2^8$ ordered coprime factorizations $ab=20!.$ Since $a\neq b,$ exactly half have $0<\frac{a}{b}<1.$ The number sought is $2^7=128.$