Let $O$ be the origin, $\overline{OA_i}=a_i$, $|a_i|=1$, $\overline{OO_i}=p_i$, $\overline{OG}=(\sum a_i)/(n+1)=:a$. We want to prove that $\sum p_i=0$. We know that $(p_i-a)^2=(p_i-a_j)^2$ for $j\ne i$.
That is, $(p_i,a_j)=(p_i,a)+a_j^2/2-a^2/2$ for $j\ne i$. Sum up by all $j\ne i$, we get $(p_i,(n+1)a-a_i)=n(p_i,a)+n/2-na^2/2$. Thus $(p_i,a_i)=(p_i,a)-n/2+na^2/2$. Therefore $(\sum_i p_i,a_j)=(p_j,a_j)+\sum_{k\ne i}(p_k,a_j)=(\sum_i p_i,a)$, $\sum_i p_i$ is orthogonal to $a-a_i$, but these vectors generate the whole space and we conclude that $\sum_i p_i=0$.