Japan Mathematical Olympiad

$3\cdot 2^{2021}+3$

For integers $n=0,\dots ,2021$, let $a\le b\le c$ be three numbers on the blackboard after the procedure is done $n$ times and let $p_n=c-b$, $q_n=b-a$. Here $n=0$ means the initial status. When the procedure is done once, three numbers $a\le b\le c$ are replaced with $\frac{a+b}{2}\le \frac{a+c}{2}\le \frac{b+c}{2}$ hence $p_{n+1}=\frac{b-a}{2}=\frac{q_n}{2}$, $q_{n+1}=\frac{c-b}{2}=\frac{p_n}{2}$ holds. Therefore $p_{2021}=\frac{q_0}{2^{2021}}$, $q_{2021}=\frac{p_0}{2^{2021}}$. Initial three numbers are distinct thus neither of $p_0,q_0$ is $0$ and then neither of $p_{2021},q_{2021}$ is $0$. All three numbers after $2021$ procedures are positive integers hence $p_{2021}\ge 1$, $q_{2021}\ge 1$ and $p_0\ge 2^{2021}$, $q_0\ge 2^{2021}$. The minimum number of initial three numbers is greater than or equal to $1$ thus the sum of initial three numbers is greater than or equal to $1+(1+2^{2021})+(1+2\cdot 2^{2021})=3\cdot 2^{2021}+3$.

We show $1$, $1+2^{2021}$, $1+2\cdot 2^{2021}$ satisfy the condition of initial three numbers. Discussion above shows $p_{2021}=1$, $q_{2021}=1$. The sum of three numbers is kept unchanged by the procedure thus the sum is $3\cdot 2^{2021}+3$ after $2021$ procedures. Therefore three numbers after $2021$ procedures are $2^{2021}$, $2^{2021}+1$, $2^{2021}+2$ and these are all integers. Hence the answer is $3\cdot 2^{2021}+3$.