with t_rank as
(
select *,
count(score) over(partition by job) as cnt,
row_number() over(partition by job order by score,id) as a,
row_number() over(partition by job order by score desc,id desc) as b
from grade
)

select id,job,score,b
from t_rank
where a >= cnt / 2 and b >= cnt / 2
order by id

中位数就是正序逆序排列都位于中间的数，两次排序后去找中间的位置