China Mathematical Competition (Complementary Test)

For $1 \le k \le n-1$, we have $0 < \sum_{i=1}^k a_i \le k$ and $0 < \sum_{i=k+1}^n a_i \le n-k$. By using the fact that $|x-y| < \max\{x, y\}$ for $x, y > 0$, we get
$$\begin{align*} |A_n - A_k| &= \left| \left(\frac{1}{n} - \frac{1}{k}\right) \sum_{i=1}^{k} a_i + \frac{1}{n} \sum_{i=k+1}^{n} a_i \right| \\ &= \left| \frac{1}{n} \sum_{i=k+1}^{n} a_i - \left(\frac{1}{k} - \frac{1}{n}\right) \sum_{i=1}^{k} a_i \right| \\ &< \max \left\{ \frac{1}{n} \sum_{i=k+1}^{n} a_i, \left(\frac{1}{k} - \frac{1}{n}\right) \sum_{i=1}^{k} a_i \right\} \\ &\le \max \left\{ \frac{1}{n}(n-k), \left(\frac{1}{k} - \frac{1}{n}\right) k \right\} \\ &= 1 - \frac{k}{n}. \end{align*}$$
Therefore,
$$\begin{align*} \left| \sum_{k=1}^{n} a_k - \sum_{k=1}^{n} A_k \right| &= \left| nA_n - \sum_{k=1}^{n} A_k \right| \\ &= \left| \sum_{k=1}^{n-1} (A_n - A_k) \right| \le \sum_{k=1}^{n-1} \left| A_n - A_k \right| \\ &< \sum_{k=1}^{n-1} \left(1 - \frac{k}{n}\right) = \frac{n-1}{2}. \end{align*}$$
This completes the proof. ☐