SQL not a single-group group function

Well the problem simply-put is that the SUM(TIME) for a specific SSN on your query is a single value, so it's objecting to MAX as it makes no sense (The maximum of a single value is meaningless).

Not sure what SQL database server you're using but I suspect you want a query more like this (Written with a MSSQL background - may need some translating to the sql server you're using):

SELECT TOP 1 SSN, SUM(TIME)FROM downloadsGROUP BY SSNORDER BY 2 DESC

This will give you the SSN with the highest total time and the total time for it.

Edit - If you have multiple with an equal time and want them all you would use:

SELECTSSN, SUM(TIME)FROM downloadsGROUP BY SSNHAVING SUM(TIME)=(SELECT MAX(SUM(TIME)) FROM downloads GROUP BY SSN))

If you want downloads number for each customer, use:

select ssn     , sum(time)  from downloads group by ssn

If you want just one record -- for a customer with highest number of downloads -- use:

select *  from (        select ssn             , sum(time)          from downloads         group by ssn         order by sum(time) desc       ) where rownum = 1

However if you want to see all customers with the same number of downloads, which share the highest position, use:

select *  from (        select ssn             , sum(time)             , dense_rank() over (order by sum(time) desc) r          from downloads         group by ssn       ) where r = 1

Maybe you find this simpler

select * from (    select ssn, sum(time) from downloads    group by ssn    order by sum(time) desc) where rownum <= 10 --top 10 downloaders

Regards
K