How to remove a specific line from a file only if there are other lines in the file

With GNU sed:

sed '0,/^AAA$/b;//d'

That is, let everything through (b branches off like a continue) up to the first AAA (from the 0th line (that is even before the first line) and the first one matching /^AAA$/ (which could be the first line)), and then for the remaining lines, delete every occurrence of AAA (an empty // pattern reuses the last pattern).

GNU sed is needed for the 0 address (and the ability to have other commands after the b one in the same expression, though that could be easily worked around in other implementations by using two -e expressions)

With awk:

awk '$0 != "AAA" || !n++'

(or for a regexp pattern: awk '!/^AAA$/ || !n++')

a shorthand for:

awk '! (&0 == "AAA" && count > 0) {print; count++}'

You can use awk as well:

awk 'NR==FNR { nums[$0]; next } !(FNR in nums)' linenum infile

in specific case when 'linenum' file might empty, awk will skip it, so it won't print whole 'infile' lines then, to fix that, use below command:

awk 'NR==FNR && FILENAME==ARGV[1]{ nums[$0]; next } !(FNR in nums)' linenum infile

or even better (thanks to Stéphane Chazelas):

awk '!firstfile_proceed { nums[$0]; next } 
     !(FNR in nums)' linenum firstfile_proceed=1 infile