select user.id,user.name,f.s_n from user,
(select user_id, sum(real_value) s_n from 
(select user_id ,if(type='reduce',-grade_num,grade_num) real_value from grade_info) g 
group by user_id) f where user.id=f.user_id and f.s_n=

(select sum(real_value) s_n from 
(select user_id ,if(type='reduce',-grade_num,grade_num) real_value from grade_info) g 
group by user_id order by s_n desc limit 1)

order by f.user_id 

思路：

1.将type中reduce和add进行分类，并改变grade_num的值（add变为正，reduce变为负）

select user_id ,if(type='reduce',-grade_num,grade_num) real_value from grade_info) g 
group by user_id

2.求出grade——info表中中各个user_id的sum

select user_id, sum(real_value) s_n from 
(select user_id ,if(type='reduce',-grade_num,grade_num) real_value from grade_info) g 
group by user_id

3.求出最大值，在第二部基础上对sum值进行降序排序，并取第一个值

select sum(real_value) s_n from 
(select user_id ,if(type='reduce',-grade_num,grade_num) real_value from grade_info) g 
group by user_id order by s_n desc limit 1

4.通过where条件将两个表合并，并求出sum值等于第三步求出的最大值即可

select user.id,user.name,f.s_n from user,
(select user_id, sum(real_value) s_n from 
(select user_id ,if(type='reduce',-grade_num,grade_num) real_value from grade_info) g 
group by user_id) f where user.id=f.user_id and f.s_n=

(select sum(real_value) s_n from 
(select user_id ,if(type='reduce',-grade_num,grade_num) real_value from grade_info) g 
group by user_id order by s_n desc limit 1)

order by f.user_id 

其他思路，可以通过窗口函数对sum值进行排序