Japan Mathematical Olympiad

85008 is the minimum value of the difference.

First we show the difference between maximum and minimum is greater than or equal to 85008. Let $a_1,a_2,\dots ,a_{2021}$ be 2021 integers satisfying the condition. Let $n$ be an integer satisfying $1\le n\le 2016$ then the condition shows $a_{n+5}-a_{n+3}>a_{n+2}-a_n$ and both left and right hand sides are integers thus $$a_{n+5}-a_{n+3}\ge a_{n+2}-a_n+1$$ holds. Let $m$ be an integer satisfying $1\le m\le 1006$, then by summing up this inequality for 506 integers $n=m,m+2,\dots ,m+1010$ we get $a_{m+1015}-a_{m+3}\ge a_{m+1012}-a_m+506$ hence $$a_{m+1015}-a_{m+1012}\ge a_{m+3}-a_m+506$$ is obtained. Then by summing up this inequality for 336 integers $m=1,4,\dots ,1006$ we get $a_{2021}-a_{1013}\ge a_{1009}-a_1+170016$ hence $(a_{2021}-a_{1013})+(a_1-a_{1009})\ge 170016$ holds. Therefore at least one of $a_{2021}-a_{1013}$ and $a_1-a_{1009}$ is greater than or equal to $\frac{170016}{2}=85008$ and thus the difference between maximum and minimum is greater than or equal to 85008.

Next we show there exists 2021 integers $a_1,a_2,\dots ,a_{2021}$ satisfying the condition whose maximum and minimum have the 85008 difference. Let $n$ be an integer satisfying $1\le n\le 2021$ and take non-negative integer $q$ less than or equal to 336 and non-negative integer $r$ less than or equal to 5 such that $n=6q+r$ holds. Define $a_n$ as $$\{\begin{array}{ll}3(q-168)(q-169)+1& (r=0),\\ (q-168)(3q+r-507)& (1\le r\le 5).\end{array}$$ If $n$ is less than or equal to 2020, $a_{n+1}-a_n$ is equal to $$\{\begin{array}{ll}(q-168)(3q+1-507)-(3(q-168)(q-169)+1)=q-169& (r=0),\\ (q-168)(3q+(r+1)-507)-(q-168)(3q+r-507)=q-168& (1\le r\le 4),\\ 3((q+1)-168)((q+1)-169)+1-(q-168)(3q+5-507)=q-167& (r=5)\end{array}$$ hence we get $$a_1\ge a_2\ge \cdots \ge a_{1010}\ge a_{1011}\le a_{1012}\le \cdots \le a_{2020}\le a_{2021}.$$ Therefore the minimum and maximum of $a_1,a_2,\dots ,a_{2021}$ are $a_{1011}=0$ and $a_1=a_{2021}=85008$ respectively then the difference is 85008. We prove those 2021 integers $a_1,a_2,\dots ,a_{2021}$ satisfy the condition. If $n$ is less than or equal to 2019, $a_{n+2}-a_n=(a_{n+2}-a_{n+1})+(a_{n+1}-a_n)$ is equal to $$\{\begin{array}{ll}2q-337& (r=0),\\ 2q-336& (1\le r\le 3),\\ 2q-335& (4\le r\le 5)\end{array}$$

and thus for $n$ less than or equal to 2016, $(a_{n+5}+a_n)-(a_{n+2}+a_{n+3})=(a_{(n+3)+2}-a_{n+3})-(a_{n+2}-a_n)$ is equal to $$\{\begin{array}{ll}(2q-336)-(2q-337)=1& (r=0),\\ (2q-335)-(2q-336)=1& (r=1,2),\\ (2(q+1)-337)-(2q-336)=1& (r=3),\\ (2(q+1)-336)-(2q-335)=1& (r=4,5)\end{array}$$ which is positive hence the condition is satisfied.

We have proved the answer is 85008.