Taiwan IMO Selection Camp, 2022

Solution. Consider
$$a_m = \begin{cases} 1, & \text{the } t\text{-th student is a girl, where } t \equiv m \pmod{2n+1}, \\ 0, & \text{otherwise.} \end{cases}$$
Consider $b_i = a_i + a_{i-k-1} - 1 \in \{-1, 0, 1\}$. It is easy to see that for all $m$,
$$b_{m+1} + \cdots + b_{m+2n+1} = 2(a_1 + \cdots + a_{2n+1}) - (2n+1) = 2(n+1) - (2n+1) = 1. \quad (1)$$
Note that this means there exists $i_0$ such that $b_{i_0} = 1$. If we can find $i$ satisfying
$$b_i = 1 \quad \text{and} \quad b_{i+1} + b_{i+2} + \cdots + b_{i+k} \ge 0, \qquad (2)$$
then $a_i = 1$ and
$$(a_{i-k} + \cdots + a_{i-1}) + (a_{i+1} + \cdots + a_{i+k}) \ge k,$$
and thus the original problem is proved.
We use proof by contradiction to show that an $i$ satisfying (2) exists. Suppose that for all $i$ satisfying $b_i = 1$, $b_{i+1} + b_{i+2} + \cdots + b_{i+k}$ is always negative. Take any one of these $i_0$, and for all $j$, define $i_j$ to be the smallest integer satisfying $i_j > i_{j-1} + k$ and $b_{i_j} = 1$. Take any two among $\{i_0, i_1, \cdots, i_{2n+1}\}$ that are congruent modulo $2n+1$, and without loss of generality assume these are $i_0$ and $i_T$.
Now, note that for all $0 \le j \le T-1$, by the definition of $i_j$, $b_{i_j + k + 1}$ through $b_{i_{j+1} - 1}$ must all be $\le 0$. By the assumption for contradiction, this means
$$S_j := b_{i_j} + \cdots + b_{i_{j+1}-1} \le b_{i_j} + \cdots + b_{i_j+k} = b_{i_j} + (b_{i_{j+1}} + \cdots + b_{i_j+k}) < 1 + 0 = 1,$$
and thus $S_j \le 0$. But on the other hand, since $(2n+1)|(i_T - i_0)$, (1) implies that
$$S_0 + \cdots + S_{T-1} = \sum_{i=i_0}^{i_T-1} b_i = \frac{i_T - i_0}{2n+1} > 0,$$
a contradiction! Hence there exists $i$ satisfying (2). This completes the proof.