这看起来像是一个缺口和孤岛问题。如果我正确地遵循了这一点，我们可以通过行号之间的差异来识别“相邻”行，然后聚合启用了标志的岛：

select min(row) as row, name, flag, sum(days) as days
from (
    select t.*,
        row_number() over(partition by name order by row) rn1,
        row_number() over(partition by name, flag order by row) rn2
    from mytable t
) t
group by name, flag, 
    case when flag = 1 then rn1 - rn2 else row end
order by 1

Demo on DB Fiddle（感谢nbk创建用例）。

你可以使用窗口函数
首先，我们需要一个分组依据的组，这两个CTE都将提供

WITH CTE As(
SELECT
     Row, Name, Flag, Days
  , CASE WHEN Flag <> LAG(Flag) OVER(PARTITION BY Name ORDEr By row)
  OR LAG(Flag) OVER(PARTITION BY Name ORDEr By row) IS NULL THEN 1 ELSe 0 ENd rnk
FROM tab1),
CTE2 AS (SELECT
Row, Name, Flag, Days, SUM(rnk) OVER(ORDER BY row) rnk
FROM CTE)
SELECT 
    Row, Name, Flag, Days
FROM CTE2 WHERE Flag = 0
UNION ALL
SELECT 
    MIN(Row), Name, Flag, SUM(Days) as Days
FROM CTE2 WHERE Flag = 1
GROUP BY Name,Flag, rnk
ORDER By row

fiddle