Target

As all of the target numbers are odd, we will need to take an even number of shots to get an even sum, in this case $80$ . Since $6$ is the only even-valued option, it just remains to show that we can get a sum of $80$ with $6$ shots. As this can be done with 1 shot of $23$ , 1 of $13$ and 4 of $11$ , we can conclude that $\boxed{6}$ is the desired answer.

Note that we could also achieve $80$ with $4$ shots, which can be done in 2 ways: (i) 3 of $23$ and 1 of $11$ , and (ii) 2 of $23$ , 1 of $21$ and 1 of $13$ .

All targets have an odd integer so if the total score is 80 that mean the number of shots is even which is 6