试试这个（Postgres 9.4+）：

select person_id, coalesce(min(days) filter (where days > 0), max(days))
from a_table
group by 1
order by 1;

Oracle设置：

CREATE TABLE table_name ( Person_Id, Days ) AS 
SELECT 1000, 100 FROM DUAL UNION ALL
SELECT 1000, 200 FROM DUAL UNION ALL
SELECT 1000, -50 FROM DUAL UNION ALL
SELECT 1000, -10 FROM DUAL UNION ALL
SELECT 1001, 100 FROM DUAL UNION ALL
SELECT 1001, 200 FROM DUAL UNION ALL
SELECT 1001,  50 FROM DUAL UNION ALL
SELECT 1001,  10 FROM DUAL UNION ALL
SELECT 1002, -50 FROM DUAL UNION ALL
SELECT 1002, -10 FROM DUAL;

查询：

SELECT person_id, days
FROM   (
  SELECT t.*,
         ROW_NUMBER() OVER ( PARTITION BY person_id
                             ORDER BY SIGN( ABS( days ) ),
                                      SIGN( DAYS ) DESC,
                                      ABS( DAYS )
                           ) AS rn
  FROM   table_name t
)
WHERE  rn = 1;

输出：

PERSON_ID       DAYS
---------- ----------
      1000        100 
      1001         10 
      1002        -10

Oracle解决方案：

with
     input_data ( person_id, days) as (
     select 1000, 100 from dual union all
     select 1000, 200 from dual union all
     select 1000, -50 from dual union all
     select 1000, -10 from dual union all
     select 1001, 100 from dual union all
     select 1001, 200 from dual union all
     select 1001,  50 from dual union all
     select 1001,  10 from dual union all
     select 1002, -50 from dual union all
     select 1002, -10 from dual
     )
select person_id,
       NVL(min(case when days > 0 then days end), max(days)) as days
from   input_data
group by person_id;


 PERSON_ID       DAYS
---------- ----------
      1000        100
      1001         10
      1002        -10

对于每个person_id，如果至少有一个days值是严格正的，那么min将只接受正的days，并将由NVL()返回。否则，min()将返回null，NVL()将在所有days上返回max()（在本例中，所有days都是负数或0）。

select Person_id, min(abs(days)) * days/abs(days) from table_name 
group by Person_id

-- + handle zero_divide ..对不起，以上内容只适用于MySQL。
类似这样的东西在任何地方都可以工作，相当于上面的查询：

select t.Person_id , min(t.days) from table_name t, 
    (select Person_id, min(abs(days)) as days from table_name group by Person_id) v 
  where t.Person_id = v.Person_id 
  and abs(t days)   = v.days 
  group by Person_id;

或

select id, min(Days) from ( 
    select Person_id, min(abs(Days)) as Days from temp group by Person_id 
    union 
    select Person_id, max(Days) as Days from temp group by Person_id
) temp 
group by Person_id;

可以通过在sql server中使用GroupBy子句来执行此操作。看看下面的查询：-

CREATE TABLE #test(Person_Id INT, [Days] INT)
DECLARE @LargestNumberFromTable INT;

INSERT INTO #test
SELECT 1000    , 100  UNION
SELECT 1000    ,  200 UNION
SELECT 1000    ,  -50 UNION
SELECT 1000    ,  -10 UNION
SELECT 1001    ,  100 UNION
SELECT 1001    ,  200 UNION
SELECT 1001    ,   50 UNION
SELECT 1001    ,   10 UNION
SELECT 1002    ,  -50 UNION
SELECT 1002    ,  -10 

SELECT @LargestNumberFromTable = ISNULL(MAX([Days]), 0)
FROM #test

SELECT Person_Id
    ,CASE WHEN  SUM(IIF([Days] > 0,[Days] , 0)) = 0 THEN MAX([Days]) -- All Negative
        WHEN SUM([Days]) = SUM(IIF([Days] > 0, [Days], 0)) THEN MIN ([Days]) -- ALL Positive
        WHEN SUM([Days]) <> SUM(IIF([Days] > 0, [Days], 0)) THEN MIN(IIF([Days] > 0, [Days], @LargestNumberFromTable)) --Mix (Negative And positive)
    END AS [Days]       
FROM #test
GROUP BY Person_Id

DROP TABLE #test