either $ 3 \nmid a_k$ or $3 \mid a_k$ the only quadratic residue $\pmod{3}$ is 1. 2012-29= 1983, which is divisible by 3

Since $29$ of them are divisible by $3$, then the other 1983 are either $1 \pmod{3}$ or $2 \pmod{3}$ Also, the $29$ numbers are also divisible by $3$ when squared $n \equiv 1 \pmod{3} \Rightarrow n^2 \equiv 1 \pmod 3$ $n \equiv 2 \pmod{3} \Rightarrow n^2 \equiv 1 \pmod 3$ Therefore all 1983 of the other numbers are of the form $1 \pmod{3}$ when squared Since $3|1983$, the sum is divisible by 3 $\blacksquare$