Invert match in sed to print all lines until line break

This would be easier to do with awk than with sed:

$ awk -v RS='' -F '\n' '!($1 ~ /ZZZ$/)' file sometext tobe here xxx_yyyy: AAA sjhdajsh skdjah hjk 

This first sets the record separator, RS, to the empty string. With this special value of RS, awk will treat each paragraph, delimited by one or more empty line, as a record. With -F '\n' we tell awk to treat each line of the paragraph as a field.

The test !($1 ~ /ZZZ$/) will be true for each record (paragraph) whose first field (1st line) does not end with the string ZZZ. Each such paragraph will be printed.

If you want an empty line after each outputted section, add -v ORS='\n\n' to the awk command line (before the awk code).

With sed, you could possibly do

$ sed '/ZZZ$/,/^$/d' file sometext tobe here xxx_yyyy: AAA sjhdajsh skdjah hjk 

I.e., "delete all lines between any line ending with ZZZ to the next empty line". You could also use the sed expression /ZZZ$/,/^$/!p with sed -n. Note however that an ZZZ on any other line other than the first line of a paragraph would also trigger this deletion.

This issue could obviously be fixed (by reading paragraphs, line by line, into the hold space, and then deciding which ones to print and which ones to discard), but it would turn into more of a bother than just using the awk variation above.

Just to comment on the expression [^ZZZ] that you use in the question. This is exactly equivalent to [^Z] and will match any single character that is not a Z. Likewise, [^text] is the same as [^etx] and would match any single character that is not an e, t, or x.