Use Complementary counting

We count the ones where the product is not divisible by $4$. Case I. The product is odd. Then, all the elements of the subset must be odd. There are $\binom{10}3=120$ ways to do this. Case II. The product is even. Then, exactly one of the elements must be $2\pmod4$, and the rest is odd. There are $5\times\binom{10}2=225$ ways to do this. In total, there are $\binom{20}3=1140$ possible susets. The answer is $1140-(120+225)=\boxed{795}$.