The prime factorization of $ 1200 $ is $ 1200=2^4 \cdot 3 \cdot 5^2 $. Thus, the factors of $ 1200 $ that can be written in the form $ n^2 $ are $ 1^2 $, $ 2^2 $, $ 4^2 $, $ 5^2 $, $ 10^2 $ and $ 20^2 $. The sum of these values of $ n $ is $ 1+2+4+5+10+20=\boxed{42} $.