The 35th Japanese Mathematical Olympiad

Set $a_{2n+1}=a_1$. Since $a_{n+1}-a_{2n+1}=-(a_1-a_{n+1})$, the values $$a_1-a_{n+1},a_2-a_{n+2},\dots ,a_n-a_{2n},a_{n+1}-a_{2n+1}$$ are all nonzero, and $a_1-a_{n+1}$ and $a_{n+1}-a_{2n+1}$ have opposite signs. Therefore, there exists an integer $m$ with $1\le m\le n$ such that $a_m-a_{n+m}$ and $a_{m+1}-a_{n+m+1}$ have opposite signs. In this case, since both $|a_m-a_{n+m}|$ and $|a_{m+1}-a_{n+m+1}|$ are at least 1, we have $$|(a_m-a_{m+1})-(a_{n+m}-a_{n+m+1})|=|(a_m-a_{n+m})-(a_{m+1}-a_{n+m+1})|\ge 2.$$ In general, for real numbers $x$ and $y$, we have $$x^2+y^2-\frac{(x-y{)}^2}{2}=\frac{(x+y{)}^2}{2}\ge 0,$$ so it follows that $$\begin{gathered} (a_m-a_{m+1}{)}^2+(a_{n+m}-a_{n+m+1}{)}^2\ge \frac{((a_m-a_{m+1})-(a_{n+m}-a_{n+m+1}){)}^2}{2} \\ \ge 2. \end{gathered}$$ Therefore, we obtain $$(a_1-a_2{)}^2+(a_2-a_3{)}^2+\cdots +(a_{2n-1}-a_{2n}{)}^2+(a_{2n}-a_1{)}^2\ge 2.$$ On the other hand, the values $a_1=a_2=\cdots =a_n=1$ and $a_{n+1}=a_{n+2}=\cdots =a_{2n}=0$ satisfy the given condition, and the value of the given function is 2 in this case. Hence, the minimum possible value is 2.