在bigdata（>1亿行）中是无法解决的

SELECT
    d[1] AS s,
    d[-1] AS e,
    arraySum(c) AS sm
FROM
(
    SELECT
        arraySplit((x, y) -> (NOT y), d, n) AS dd,
        arraySplit((x, y) -> (NOT y), c, n) AS cc
    FROM
    (
        SELECT
            groupArray(date) AS d,
            groupArray(ne) AS n,
            groupArray(change) AS c
        FROM
        (
            SELECT *
            FROM mytable
            ORDER BY rn ASC
        )
    )
)
ARRAY JOIN
    dd AS d,
    cc AS c

┌─s──────────┬─e──────────┬───────────sm─┐
│ 2008-12-07 │ 2009-04-26 │   -725166328 │
│ 2013-07-06 │ 2015-06-08 │ -15224653176 │
└────────────┴────────────┴──────────────┘

这是一个缺口和孤岛问题。规范的解决方案确实使用了窗口函数，而clickhouse不支持这些函数。
下面是一种使用子查询模拟条件窗口和的方法：

select
    min(date) start_date,
    max(date) end_date,
    sum(change) sum_change
from (
    select 
        t.*,
        (select count(*) from mytable t1 where t1.date <= t.date and t1.ne = 0) grp
    from mytable t
) t
group by grp

子查询统计有多少行 ne = 0 从表的第一行到当前行。这定义了记录组。那么剩下要做的就是聚合。
如果您可以使用窗口函数，您可以将其表述为：

select
    min(date) start_date,
    max(date) end_date,
    sum(change) sum_change
from (
    select 
        t.*,
        sum(case when ne = 0 then 1 else 0 end) over(order by date) grp
    from mytable t
) t
group by grp

只是解决此任务的另一种方法：

WITH (SELECT arraySort(groupArray(rn))
    FROM test_table
    WHERE ne = 0) as group_start_id
SELECT argMin(date, rn) start, argMax(date, rn) end, sum(change)
FROM (
    SELECT rn, date, change
    FROM test_table
    ORDER BY rn)
GROUP BY arrayFirstIndex(x -> rn < x, group_start_id)   
ORDER BY start

样本数据测试：

WITH (SELECT arraySort(groupArray(rn))
    FROM (
        SELECT data.1 rn, data.2 date, data.3 ne, data.4 change
        FROM (
            SELECT arrayJoin([
            (0, toDate('2008-12-07'), 0, toInt64(-10330848398)),
            (1, toDate('2009-04-14'), 1, toInt64(-61290)),
            (2, toDate('2009-04-26'), 1, toInt64(9605743360)),
            (3, toDate('2013-07-06'), 0, toInt64(-32028871920)),
            (4, toDate('2014-01-12'), 1, toInt64(-42296164902)),
            (5, toDate('2015-06-08'), 1, toInt64(59100383646)),
            (6, toDate('2015-06-08'), 0, toInt64(101)),
            (7, toDate('2015-06-09'), 0, toInt64(102)),
            (8, toDate('2015-06-10'), 0, toInt64(103)),
            (9, toDate('2015-06-11'), 1, toInt64(104))
            ]) data))
    WHERE ne = 0) as group_start_id
SELECT argMin(date, rn) start, argMax(date, rn) end, sum(change)
FROM (
    SELECT data.1 rn, data.2 date, data.4 change
    FROM (
        SELECT arrayJoin([
        (0, toDate('2008-12-07'), 0, toInt64(-10330848398)),
        (1, toDate('2009-04-14'), 1, toInt64(-61290)),
        (2, toDate('2009-04-26'), 1, toInt64(9605743360)),
        (3, toDate('2013-07-06'), 0, toInt64(-32028871920)),
        (4, toDate('2014-01-12'), 1, toInt64(-42296164902)),
        (5, toDate('2015-06-08'), 1, toInt64(59100383646)),
        (6, toDate('2015-06-08'), 0, toInt64(101)),
        (7, toDate('2015-06-09'), 0, toInt64(102)),
        (8, toDate('2015-06-10'), 0, toInt64(103)),
        (9, toDate('2015-06-11'), 1, toInt64(104))
        ]) data)
    ORDER BY rn)
GROUP BY arrayFirstIndex(x -> rn < x, group_start_id)   
ORDER BY start
/* result
┌──────start─┬────────end─┬──sum(change)─┐
│ 2008-12-07 │ 2009-04-26 │   -725166328 │
│ 2013-07-06 │ 2015-06-08 │ -15224653176 │
│ 2015-06-08 │ 2015-06-08 │          101 │
│ 2015-06-09 │ 2015-06-09 │          102 │
│ 2015-06-10 │ 2015-06-11 │          207 │
└────────────┴────────────┴──────────────┘

* /

选择ne，min（日期）作为开始，max（日期）作为结束，sum（更改）作为更改组by ne

假设clickhouse支持变量：

set @block := -1;
select 
    block as row,
    min(date) as start,
    max(date) as end,
    sum(change)
from
    (select  
        case when ne = 0 then @block:=@block+1 end as dummy,
        @block as block,
        t.*
    from t) tt
group by block;