China National Team Selection Test

Define $I = \{k \mid b_k \ge 1\}$. For every $k \in I$, assume that the maximum value $b_k$ of $\frac{1}{2m+1} \sum_{i=k-m}^{k+m} a_i$ is attained at $m = m_k$, and call
$$[k - m_k, k + m_k] := \{k - m_k, k - m_k + 1, \dots, k + m_k\}$$
a “nice segment”, where the subscripts are taken modulo $2n+1$. Obviously, the union of all nice segments contains $I$.

Claim There exists a collection of nice segments whose union contains $I$, and moreover, each $i \in \{1, 2, \dots, 2n + 1\}$ is contained in at most two segments.

Proof of claim If $[1, 2n + 1]$ is a nice segment, the conclusion is trivial. In the following, assume this is not the case. If $i$ is contained in $r$ nice segments
$$[i - u_1, i + v_1], \dots, [i - u_r, i + v_r]$$
where $r \ge 3$ and $0 \le u_j, v_j < 2n$, let $u_j = \max\{u_1, \dots, u_r\}$ and $v_k = \max\{v_1, \dots, v_r\}$. We can keep the nice segments $[i - u_j, i + v_j]$, $[i - u_k, i + v_k]$ and drop the other $r-2$ segments. Now the segments still cover $1, \dots, 2n+1$, and at most two of them cover $i$. For each $i \in \{1, 2, \dots, 2n+1\}$, perform the above operation. Eventually, we find a collection of nice segments with the desired properties.

For the original problem, let $[i_1 - m_1, i_1 + m_1]$, $\dots$, $[i_r - m_r, i_r + m_r]$ be a collection of nice segments chosen in the claim. We have
$$2 \sum_{i=1}^{2n+1} a_i \ge \sum_{\alpha=1}^{r} \sum_{k=i_{\alpha}-m_{\alpha}}^{i_{\alpha}+m_{\alpha}} a_k = \sum_{\alpha=1}^{r} (2m_{\alpha} + 1)b_{i_{\alpha}} \ge \sum_{\alpha=1}^{r} (2m_{\alpha} + 1) \ge |I|,$$
the first inequality is due to each $i$ being contained in at most two nice segments; the equality and the next inequality are due to the definition of nice segments; the last inequality is due to the union of the nice segments containing $I$. $\square$