Find and sum date ranges with overlapping records in postgresql

demo:db<>fiddle (uses the old data set with the overlapping A-B-part)

Disclaimer: This works for day intervals not for timestamps. The requirement for ts came later.

SELECT
    s.acts,
    s.sum,
    MIN(a.start) as start,
    MAX(a.end) as end
FROM (
    SELECT DISTINCT ON (acts)
        array_agg(name) as acts,
        SUM(count)
    FROM
        activities, generate_series(start, "end", interval '1 day') gs
    GROUP BY gs
    HAVING cardinality(array_agg(name)) > 1
) s
JOIN activities a
ON a.name = ANY(s.acts)
GROUP BY s.acts, s.sum

1. generate_series generates all dates between start and end. So every date an activity exists gets one row with the specific count
2. Grouping all dates, aggregating all existing activities and sum of their counts
3. HAVING filters out the dates where only one activity exist
4. Because there are different days with the same activities we only need one representant: Filter all duplicates with DISTINCT ON
5. Join this result against the original table to get the start and end. (note that "end" is a reserved word in Postgres, you should better find another column name!). It was more comfortable to lose them before but its possible to get these data within the subquery.
6. Group this join to get the most early and latest date of each interval.

---

Here's a version for timestamps:

demo:db<>fiddle

WITH timeslots AS (
    SELECT * FROM (
        SELECT
            tsrange(timepoint, lead(timepoint) OVER (ORDER BY timepoint)),
            lead(timepoint) OVER (ORDER BY timepoint)     -- 2
        FROM (
            SELECT 
                unnest(ARRAY[start, "end"]) as timepoint  -- 1 
            FROM
                activities
            ORDER BY timepoint
        ) s
    )s  WHERE lead IS NOT NULL                            -- 3
)
SELECT 
    GREATEST(MAX(start), lower(tsrange)),                 -- 6
    LEAST(MIN("end"), upper(tsrange)),
    array_agg(name),                                      -- 5
    sum(count)
FROM 
    timeslots t
JOIN activities a
ON t.tsrange && tsrange(a.start, a.end)                   -- 4
GROUP BY tsrange
HAVING cardinality(array_agg(name)) > 1

The main idea is to identify possible time slots. So I take every known time (both start and end) and put them into a sorted list. So I can take the first tow known times (17:00 from start A and 18:00 from start B) and check which interval is in it. Then I check it for the 2nd and 3rd, then for 3rd an 4th and so on.

In the first timeslot only A fits. In the second from 18-19 also B is fitting. In the next slot 19-20 also C, from 20 to 20:30 A isn't fitting anymore, only B and C. The next one is 20:30-22 where only B fits, finally 22-23 D is added to B and last but not least only D fits into 23-23:30.

So I take this time list and join it agains the activities table where the intervals intersect. After that its only a grouping by time slot and sum up your count.

1. this puts both ts of a row into one array whose elements are expanded into one row per element with unnest. So I get all times into one column which can be simply ordered
2. using the lead window function allows to take the value of the next row into the current one. So I can create a timestamp range out of these both values with tsrange
3. This filter is necessary because the last row has no "next value". This creates a NULL value which is interpreted by tsrange as infinity. So this would create an incredible wrong time slot. So we need to filter this row out.
4. Join the time slots against the original table. The && operator checks if two range types overlap.
5. Grouping by single time slots, aggregating the names and the count. Filter out the time slots with only one activity by using the HAVING clause
6. A little bit tricky to get the right start and end points. So the start points are either the maximum of the activity start or the beginning of a time slot (which can be get using lower). E.g. Take the 20-20:30 slot: It begins 20h but neither B nor C has its starting point there. Similar the end time.