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
$$x^2 + y^2 - \frac{(x - y)^2}{2} = \frac{(x + y)^2}{2} \geq 0,$$
so it follows that
$$(a_m - a_{m+1})^2 + (a_{n+m} - a_{n+m+1})^2 \ge \frac{((a_m - a_{m+1}) - (a_{n+m} - a_{n+m+1}))^2}{2} \\ \ge 2.$$
Therefore, we obtain
$$(a_1 - a_2)^2 + (a_2 - a_3)^2 + \dots + (a_{2n-1} - a_{2n})^2 + (a_{2n} - a_1)^2 \ge 2.$$
On the other hand, the values $a_1 = a_2 = \dots = a_n = 1$ and $a_{n+1} = a_{n+2} = \dots = a_{2n} = 0$ satisfy the given condition, and the value of the given function is 2 in this case. Hence, the minimum possible value is 2.