Run a cron job on the first Monday of every month?

You can put the condition into the actual crontab command (generic way):

[ "$(date '+%u')" = "1" ] && echo "It's Monday"

if your locale is EN/US, you can also compare strings (initial answer):

[ "$(date '+%a')" = "Mon" ] && echo "It's Monday"

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Now, if this condition is true on one of the first seven days in a month, you have its first Monday. Note that in the crontab, the percent-syntax needs to be escaped though (generic way):

0   12  1-7 *   *   [ "$(date '+\%u')" = "1" ] && echo "It's Monday"

if your locale is EN/US, you can also compare strings (initial answer):

0   12  1-7 *   *   [ "$(date '+\%a')" = "Mon" ] && echo "It's Monday"

Replace the echo command with the actual command you want to run. I found a similar approach too.

I have a computer with locale on Spanish, so, this approach isn't working for me because mon changes to lun

Other languages would fail as well, so, I did a slight variation on the accepted answer that takes out the language barrier:

 0 9 1-7 * *   [ "$(date '+\%u')" = "1" ] && echo "¡Es lunes!"

I find it easier when there's no need to handle day numbers.

Run First Monday of month:

0 2 * * 1 [ `date '+\%m'` == `date '+\%m' -d "1 week ago"` ] || /path/to/command

i.e. if the month 1 week ago is not the same as the current month then we are on the 1st day 1 (= Monday) of the month.

Similarly, for the Third Friday

0 2 * * 6 [ `date '+\%m'` == `date '+\%m' -d "3 weeks ago"` ] || /path/to/command

i.e. if the month 3 weeks ago is different to current month then we are on the 3rd day 6 (= Friday) of the month