China Girls' Mathematical Olympiad

Let $b_0 = 0$, $b_1 = 1$, $b_n = 4b_{n-2} + b_{n-1}$ ($n \ge 2$). Then
$$b_n = \frac{(2+\sqrt{5})^n - (2-\sqrt{5})^n}{2\sqrt{5}}.$$
In particular, $b_6 = 1292$, $b_7 = 5473$.
For every $k = 1, 2, \dots, 5473$, there are unique integers $x_k, y_k$ such that $1292k = x_k + 5473y_k$, and $1 \le x_k \le 5473$. Since $(1292, 5473) = 1$, $x_1, x_2, \dots, x_{5473}$ is a permutation of $\{1, 2, \dots, 5473\}$, and it is clear that $\{y_k\}$ is nondecreasing:
$$y_1 \le y_2 \le \dots \le y_{5473} = 1291.$$
For our convenience, let $f(x) = x - \lfloor x \rfloor$. We have
$$\begin{aligned} f(x_k\sqrt{5}) &= f(1292k\sqrt{5} - 5473y_k\sqrt{5}) \\ &= f\left(\frac{(2+\sqrt{5})^6 - (2-\sqrt{5})^6}{2}k - \frac{(2+\sqrt{5})^7 - (2-\sqrt{5})^7}{2}y_k\right) \\ &= f(-(2-\sqrt{5})^6k + (2-\sqrt{5})^7y_k). \end{aligned}$$
Since
$$\begin{aligned} & 0 < (2-\sqrt{5})^6 k - (2-\sqrt{5})^7 y_k \\ & \le 5473(2-\sqrt{5})^6 - 1291(2-\sqrt{5})^7 < 1, \end{aligned}$$
it follows that
$$f(x_k \sqrt{5}) = 1 - (2 - \sqrt{5})^k + (2 - \sqrt{5})^7 y_k,$$
and this is strictly decreasing.
Now, since $x_1 = 1292$, $x_{5473} = 5473$, $x_{5472} = 4181$, $x_{5471} = 2889$, $x_{5470} = 1597$, we see that $a_{1292}$ attains the maximum and $a_{1597}$ the minimum.