Database

select * from Orders where OrderValue > 500

Field1  Field2<br />
3	34<br />
4	58<br />
5	97<br />
6	234<br />
7	345<br />
8	540<br />
9	679<br />
10	768<br />
11	2312<br />
12	4324

select * from MyTable
  where Field1 in (
     select a.Field1
     from MyTable a
     inner join MyTable b
       on b.Field1 <= a.Field1
     group by a.Field1
     having sum(Field2) < 5000)
  order by Field1

field1	field2<br />
10       768<br />
11      2312<br />
4         58<br />
5         97<br />
7        345

create table #MyTable (
  Field1 int,
  Field2 int
  )
go

insert into #MyTable values (3, 34)
insert into #MyTable values (4, 58)
insert into #MyTable values (5, 97)
insert into #MyTable values (6, 234)
insert into #MyTable values (7, 345)
insert into #MyTable values (8, 540)
insert into #MyTable values (9, 679)
insert into #MyTable values (10, 768)
insert into #MyTable values (11, 2312)
insert into #MyTable values (12, 4324)
go

select * from #MyTable
  where Field1 in (
     select a.Field1
     from #MyTable a
     inner join #MyTable b 
       on b.Field1 <= a.Field1
     group by a.Field1
     having sum(b.Field2) < 5000)
  order by Field1
go

andyharman wrote:

My previous solution had an ambiguous column.

select Field2,sum(Field2)as [Sum] from tablename where Field2< 5000<br />
group by Field2 

Blue_Boy wrote:

select Field2,sum(Field2)as [Sum] from tablename where Field2< 5000
group by Field2

select * from tablename  where Field2 in<br />
 (select firstTable.Field2   from tablename firstTable     inner join tablename secondTable        <br />
on secondTable.Field2 <= firstTable.Field2 group by firstTable.Field2 having sum(secondTable.Field2) < 5000)  order by Field2