Lol , Units digit is a distraction. A number is divisible by $2$ iff its unit digit or last digit is divisible by $2$. Thus taking the whole sequence in $\mathbb{F}_2$ , we get $$1,1,0,1,0,1,1,0,0,1,0,0,0,1,1,1,1,0,1,0,1,1,\cdots$$Observe the sequence has a period of $15$ , Thus $a_{n} \equiv a_{n (\text{mod} 15)}(\text{mod}2)$. Now note that $2000 \equiv 5 (\text{mod} 15)$ and $1985 \equiv 5(\text{mod}15)$, Therefore $$a^2_{1985}+a^2_{1986}+…+a^2_{2000} \equiv a_{5}^2 + \cdots + a_{20}^2 \equiv 0(\text{mod}2)$$which is trivial to see.