Let $n \geq 4$ be an integer. Find all positive real solutions to the following system of $2 n$ equations:
$$
\begin{aligned}
a_{1} & =\frac{1}{a_{2 n}}+\frac{1}{a_{2}}, & a_{2} & =a_{1}+a_{3} \\
a_{3} & =\frac{1}{a_{2}}+\frac{1}{a_{4}}, & a_{4} & =a_{3}+a_{5} \\
a_{5} & =\frac{1}{a_{4}}+\frac{1}{a_{6}}, & a_{6} & =a_{5}+a_{7} \\
& \vdots & & \vdots \\
a_{2 n-1} & =\frac{1}{a_{2 n-2}}+\frac{1}{a_{2 n}}, & a_{2 n} & =a_{2 n-1}+a_{1}
\end{aligned}
$$
The answer is that the only solution is $(1,2,1,2, \ldots, 1,2)$ which works.
We will prove $a_{2 k}$ is a constant sequence, at which point the result is obvious.
First approach (Andrew Gu). Apparently, with indices modulo $2 n$, we should have
$$
a_{2 k}=\frac{1}{a_{2 k-2}}+\frac{2}{a_{2 k}}+\frac{1}{a_{2 k+2}}
$$
for every index $k$ (this eliminates all $a_{\text {odd }}$ 's). Define
$$
m=\min _{k} a_{2 k} \quad \text { and } \quad M=\max _{k} a_{2 k} .
$$

Look at the indices $i$ and $j$ achieving $m$ and $M$ to respectively get
$$
\begin{array}{l}
m=\frac{2}{m}+\frac{1}{a_{2 i-2}}+\frac{1}{a_{2 i+2}} \geq \frac{2}{m}+\frac{1}{M}+\frac{1}{M}=\frac{2}{m}+\frac{2}{M} \\
M=\frac{2}{M}+\frac{1}{a_{2 j-2}}+\frac{1}{a_{2 j+2}} \leq \frac{2}{M}+\frac{1}{m}+\frac{1}{m}=\frac{2}{m}+\frac{2}{M} .
\end{array}
$$

Together this gives $m \geq M$, so $m=M$. That means $a_{2 i}$ is constant as $i$ varies, solving the problem.
₪ Second approach (author's solution). As before, we have
$$
a_{2 k}=\frac{1}{a_{2 k-2}}+\frac{2}{a_{2 k}}+\frac{1}{a_{2 k+2}}
$$

The proof proceeds in three steps.
- Define
$$
S=\sum_{k} a_{2 k}, \quad \text { and } \quad T=\sum_{k} \frac{1}{a_{2 k}} .
$$

Summing gives $S=4 T$. On the other hand, Cauchy-Schwarz says $S \cdot T \geq n^{2}$, so $T \geq \frac{1}{2} n$.
- On the other hand,
$$
1=\frac{1}{a_{2 k-2} a_{2 k}}+\frac{2}{a_{2 k}^{2}}+\frac{1}{a_{2 k} a_{2 k+2}}
$$

Sum this modified statement to obtain
$$
n=\sum_{k}\left(\frac{1}{a_{2 k}}+\frac{1}{a_{2 k+2}}\right)^{2} \stackrel{\text { QM-AM }}{\geq} \frac{1}{n}\left(\sum_{k} \frac{1}{a_{2 k}}+\frac{1}{a_{2 k+2}}\right)^{2}=\frac{1}{n}(2 T)^{2}
$$

So $T \leq \frac{1}{2} n$.
- Since $T \leq \frac{1}{2} n$ and $T \geq \frac{1}{2} n$, we must have equality everywhere above. This means $a_{2 k}$ is a constant sequence.

Remark. The problem is likely intractable over $\mathbb{C}$, in the sense that one gets a high-degree polynomial which almost certainly has many complex roots. So it seems likely that most solutions must involve some sort of inequality, using the fact we are over $\mathbb{R}_{>0}$ instead.