Question 74
Initially, only the integer $44$ is written on a board. An integer $a$ on the board can be replaced with four pairwise different integers $a_{1}, a_{2}, a_{3}, a_{4}$ such that the arithmetic mean $\frac{1}{4}\left(a_{1}+a_{2}+a_{3}+a_{4}\right)$ of the four new integers is equal to the number $a$. In a step we simultaneously replace all the integers on the board in the above way. After $30$ steps we end up with $n=4^{30}$ integers $b_{1}, b_{2}, \ldots, b_{n}$ on the board. Prove that
$$
\frac{b_{1}^{2}+b_{2}^{2}+\cdots+b_{n}^{2}}{n} \geqslant 2011
$$
$$
\frac{b_{1}^{2}+b_{2}^{2}+\cdots+b_{n}^{2}}{n} \geqslant 2011
$$