-- Тестовые данные:
with 
  t(id, dated, summa) as 
    (
      select * 
        from (
               values 
                      (1, 20120512, 536),
                      (1, 20120426, 450),
                      (1, 20120412, 350),
                      (1, 20120401, 0),
                      (1, 20120321, 100),
                      (1, 20120301, 0),
                      (1, 20120201, 50),
                      (1, 20120101, 0),
                      (1, 20110301, 5),
                      (2, 20130710, 14500),
                      (2, 20130528, 0),
                      (2, 20130402, 200),
                      (3, 20121102, 700),
                      (4, 20120823, 450),
                      (4, 20120705, 210),
                      (4, 20120411, 110),
                      (5, 20130217, 0),
                      (5, 20130109, 2000)
             )v(a,b,c)
    )
--
-- Основной запрос:
select id, min(dated) as dated, max(summa) as summa
  from (
         select t.id, t.dated,
                case when dated = max(dated) over(partition by id) then summa end as summa,
                sum(case when summa = 0 then 1 else 0 end) 
                  over(partition by id 
                           order by dated desc) as grp_id
           from t
       )v
 where grp_id = 0
 group by id;