The keys as it were are to note that \n\n is a blank line or if you read a file line by line then a blank line will match m/^\s*$/

perl -pi.bak -e 's/\n{2,}/\n/' file.txt

perl -pi -e 's/^\s*\z//' file.txt

perl -ne 'print unless m/^\s*$/' file.txt > noblanks.txt
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Update

Updated as per sgifford's comments.

A few nits: The first answer requires that Perl read in slurp mode, and needs a g modifier to replace all occurences:

perl -p -e 'BEGIN { undef $/ } s/\n{2,}/\n/g'
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Also, the two examples behave differently in the face of lines that contain only blank characters, like space and tab. The second will delete them, while the first will leave them intact. You can make the second leave them intact by using m/^$/; to make the first delete whitespace-only lines I think s/\n\s*\n/\n/g will work, but I haven't thought it through yet.

perl -p -e 'BEGIN { undef $/ } s/\n{2,}/\n/g'
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Which can be shortened to:

perl -p0777e 's/\n{2,}/\n/g'
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See perldoc perlrun for details of perl's -0 option.

It is as you say.

perl -pi -e 's/^\s*\z//' file.txt deals with blank lines with other whitespace.

Perhaps it could be useful (in case that these lines are not really empty because they contain whitespace) to use a script like this:

while (<>) {
    print unless m~^\s*\n$~;
}
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This is surely not the most elegant solution, but it should do the task.

None of the non-<>-print solutions I see here will cope with a single opening blank line since they all search for multiple \n characters. This does the trick using the /m modifier, which allows ^ to match the beginning of a line anywhere in your string:

perl -p0777e 's/^\n+//mg'