Special Order By Clause

Hello NG
As initial position the following table:
ID Name Flag <Anything> Date
1 Müller 0 .... 01.01.2002
500 Schmitt 0 .... 25.11.2002
7500 Graf 1 .... 30.11.2002
100000 Seiler 1 .... 31.12.2002
The ID is to illustrate that in the table are > 100000 records. <something> is to show that there are still other columns in the table.
A date shall be given (e.g. the 30.11.2002)
What i wish is a sql query, which returns exactly 1000 records under the condition that all records from the given date backwards are always included. That is independently wether I sort ASC or DESC the records with date = 30.11.2002 must be in the result. Even if I sort ASC or DESC according to other columns than date, all records must be contained of 30.11.2002 plus the other records until 1000 records are found.
If I try to sort with ORDER BY date ASC naturally the records with date 01.01.2002 are the first. From the records with 30.11.2002 I do not see anything because the 1000 desired records are already filled with 'old' data.
If there are more than 1000 records available with date equal to 30.11.2002 ists all the same if the result is filled with only this[i]Long postings are being truncated to ~1 kB at this time.

Thanx for your help John
Principally the query seems to work as desired. Unfortonatelly i receive the error 'ORA-0331: end-of-file on communication channel' when executing the whole query. If executing only the part after the first from then it works. Actually I have defined 30 fields in each select statement.
Then I have hocused me in the amount of records in the table. There are actually > 7 Million records!
How can I resolve this problem?
Thanx in advance
Adrian

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    <BLOCKQUOTE><font size="1" face="Verdana, Arial">quote:</font><HR>Originally posted by Vojtech Novacek:
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    A subquery in a SELECT list
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