思路:中位数的正序>2/total,倒序>2/total(即使是累加的排名)
select
grade
from
(
select
grade
,sum(number)over(order by grade) t_rank
,sum(number)over(order by grade desc) t_rankk
,sum(number)over() total
from class_grade
) a
where t_rank>=total/2 and t_rankk>=total/2
order by grade