The integers $1,2, \ldots, 7$ are written on the board. At each step, we are allowed to erase two numbers $a$ and $b$ from the board and, in their place, write $\frac{2 a+b}{3}$ and $\frac{a+2 b}{3}$.

1. After a finite number of steps, can we obtain a board on which the number 3 appears 7 times?
2. After a finite number of steps, can we obtain a board on which the number 4 appears 7 times?
1. We notice that $\frac{2 a+b}{3}+\frac{a+2 b}{3}=a+b$. The sum of the numbers written on the board is therefore an invariant equal to $1+2+\ldots+7=\frac{8 \times 7}{2}=28$. Since $28 \neq 3 \times 7=21$, it is impossible to obtain 7 numbers '3'.

2. Suppose we obtain 7 numbers '4'. Then at some step, we get 5 numbers '4' and two distinct numbers $a$ and $b$ such that $\frac{a+2 b}{3}=\frac{2 a+b}{3}$, so $a=b$. Impossible.