Easy and obvious solutions are 2018×2019×2020, 2019×2020×2021, 2020×2021×2022.
REVISION FOLLOWING COMMENT
USING QUADRATICS
We need to consider p(p+1)=2020n where n is an integer.
This gives us the quadratic p2+p-2020n=0, so p=(-1±√(8080n+1))/2.
8080n+1=x2 where x is an integer. When n=5, x2=40401=2012, so p=(-1±201)/2=-101 or 100.
So p(p+1)=100×101=10100. This gives us also p+1=100 and p+2=101, p=99.
Therefore 100×101×102 and 99×100×101 are also divisible by 2020.
When n=81, p=(-1±√654481)/2=404 or -405.
Therefore 404×405×406 and 403×404×405 are also divisible by 2020.
When n=126, p=(-1±√1018081)/2=404 or -405.
Therefore 504×505×506 and 503×504×505 are also divisible by 2020.
FACTOR PAIRS
A better way than using the quadratic is to look at the factors of 2020:
2020=22×5×101, creating factor pairs (2,1010), (4,505), (5,404), (10,202), (20,101).
If we use quadratics we also have to consider p(p+2)=2020n, p2+2p-2020n=0. So 1010×1011×1012 is also divisible by 2020. As is 200×201×202. These solutions can also be deduced from the factor pairs. That makes 11 in all.