We note that $20!$ has $8$ different prime factors. Each prime factor, together with its exponent in the prime factorization of $20!$ can be part of either the denominator or the numerator, but not both, in order to make the result irreducible. Hence, there are $2^8$ different combinations. Meanwhile, every case in which a prime factor goes to denominator must correspond to a case in which this prime factor goes to the numerator. These two cases form a bijection. In every such pair, only one resulting fraction will be less that $1$. Therefore, we conclude the answer is $$2^7=\boxed{128}$$