Select all overlapping ranges from starting range
A straightforward recursive solution splits the task in two parts:


Find the earliest start date by following the chain of overlapping ranges
Find the latest end date by following the chain of overlapping ranges


Details follow:

1. Follow the chain of overlaps leading to the earliest start date


Find the starting range
Find the overlapping range with the latest earlier start date than current
Go to step 2


Example code:

DECLARE @STARTDATE DATETIME = '2018-08-24 12:00:00';
DECLARE @ENDDATE DATETIME = '2018-08-31 12:00:00';

WITH MinStart AS
(
    -- Starting range
    SELECT TOP (1) 
        D.ID,
        D.STARTDATE
    FROM dbo.DATA1 AS D
    WHERE 
        D.STARTDATE &gt;= @STARTDATE
        AND D.ENDDATE &lt;= @ENDDATE
    ORDER BY 
        D.ID ASC

    UNION ALL

    SELECT
        P1.ID,
        P1.STARTDATE
    FROM 
    (
        -- Overlapping ranges with an earlier start date
        -- Numbered starting with the highest start date
        -- Tie-break dates using ID
        SELECT
            RN = ROW_NUMBER() OVER (
                ORDER BY D.STARTDATE DESC, D.ID ASC),
            D.ID,
            D.STARTDATE
        FROM MinStart AS R
        JOIN dbo.DATA1 AS D
            ON D.ENDDATE &gt;= R.STARTDATE
            AND D.STARTDATE &lt; R.STARTDATE
    ) AS P1
    WHERE
        -- Highest start date only
        P1.RN = 1
)
SELECT
    MS.ID,
    MS.STARTDATE
FROM MinStart AS MS
OPTION (MAXRECURSION 0);


2. Follow the chain of overlaps leading to the latest end date


Find the starting range
Find the overlapping range with the earliest later end date than current
Go to step 2


Example code:

DECLARE @STARTDATE DATETIME = '2018-08-24 12:00:00';
DECLARE @ENDDATE DATETIME = '2018-08-31 12:00:00';

WITH MaxEnd AS
(
    -- Starting range
    SELECT TOP (1) 
        D.ID, 
        D.ENDDATE
    FROM dbo.DATA1 AS D
    WHERE
        D.STARTDATE &gt;= @STARTDATE
        AND D.ENDDATE &lt;= @ENDDATE
    ORDER BY 
        D.ID ASC

    UNION ALL

    SELECT
        P1.ID,
        P1.ENDDATE
    FROM 
    (
        -- Overlapping ranges with a later end date
        -- Numbered starting with the lowest end date
        -- Tie-break dates using ID
        SELECT
            RN = ROW_NUMBER() OVER (
                ORDER BY D.ENDDATE ASC, D.ID ASC),
            D.ID,
            D.ENDDATE
        FROM MaxEnd AS R
        JOIN dbo.DATA1 AS D
            ON D.ENDDATE &gt; R.ENDDATE
            AND D.STARTDATE &lt; R.ENDDATE
    ) AS P1
    WHERE 
        -- Lowest end date only
        P1.RN = 1
)
SELECT
    ME.ID,
    ME.ENDDATE
FROM MaxEnd AS ME
OPTION (MAXRECURSION 0);


With suitable indexing, both queries use an execution plan like:



The results given the sample data provided are:


╔════╦═════════════════════════╗
║ ID ║        STARTDATE        ║
╠════╬═════════════════════════╣
║  3 ║ 2018-08-24 12:00:00.000 ║
║  6 ║ 2018-08-23 12:00:00.000 ║
║  2 ║ 2018-08-22 12:00:00.000 ║
╚════╩═════════════════════════╝
╔════╦═════════════════════════╗
║ ID ║         ENDDATE         ║
╠════╬═════════════════════════╣
║  3 ║ 2018-08-29 12:00:00.000 ║
║  6 ║ 2018-08-30 12:00:00.000 ║
║  4 ║ 2018-09-02 12:00:00.000 ║
╚════╩═════════════════════════╝


So IDs 2, 3, 4, and 6 were used. The final range is 2018-08-22 12:00:00.000 (lowest start date found) to 2018-09-02 12:00:00.000 (highest end date found).

Note: The code above includes a tie-break on dates (using the ID column) due to duplicates in the sample data. This may, or may not, be required to solve the real problem.

The indexes used were:

CREATE TABLE dbo.DATA1
(
  ID integer PRIMARY KEY,
  STARTDATE datetime NOT NULL,
  ENDDATE datetime NOT NULL,
);

CREATE UNIQUE INDEX i1 ON dbo.DATA1 (STARTDATE DESC, ID ASC) INCLUDE (ENDDATE);
CREATE UNIQUE INDEX i2 ON dbo.DATA1 (ENDDATE ASC, ID DESC) INCLUDE (STARTDATE);


Demo: db&lt;>fiddle

Note that truly optimal indexing for these types of queries can be difficult. Even so, the approach above often performs well for many practical tasks in this area.

Until proper support for interval queries and predicates is added to SQL Server, higher performance solutions require significant extra work. See:


Interval Queries in SQL Server Part 4 by Dejan Sarka
Interval Queries in SQL Server by Itzik Ben-Gan