You may wish to be more specific about "oldest file", as if you continue to move files, even one just accessed/written to a second ago will be technically the "oldest file" if it's the only one left. If you're doing something like removing old logs or backup, you might find "find"'s tests more helpful. I wouldn't use cron for this: try a "for" loop with a sleep timer. Something like (for Zsh):
for filez ( `find $SOURCE_DIR -atime +1 | xargs` ); do
echo "Moving $filez accessed at least two days ago and waiting 60 seconds..."
mv $filez $DEST_DIR
If you want the oldest file in a directory, and only that one file, then 'ls -t | tail -n 1' in that directory will get you that file.
See 'man find', specifically the section on "TESTS", if the above example needs changing to suit your needs. If you were going to perform this removal on a schedule, such as daily or weekly, then the above code could fit well into a cron job, or a script called via a cron job. How to do it will depend on which cron daemon you have. Personally, I prefer fcron.