There is a another solution using Interval Tree.

let's T[v] == 1 if number v is inside the black box, otherwise T[v] == 0

Using array T we can construct interval tree.
Each node will contain basis vectors of existing numbers in it's range.

Initially Array T is filled with 0.
Adding number v, means T[v] = 1
Deleting number v, means T[v] = 0
After each operation we need to update only log(n) nodes.

I've sent stupid solution and got almost all points (131.71)

Idea is: have a set with current basis and set of others.
Light operations:
Add, delete from others

Heavy operation:
delete from basis.(We should check if there are any element in others that we can add to basis — deleted element. It's O(others_size) but to make it a bit random (More random elements go to current basis, so it's harder to remove them and not element from others) I support an iterator to the set of others and check all element of others in order [iter, end), then [begin, iter), not [begin, end) but I am not sure if it helped at all (probably just check from begin to end will work too)

That a thing why I don't really like contests with not full solutions getting points.