Assume $1 \le a_1 < a_2 < \dots < a_m \le n$.

For even $m$ we can group the elements of $A$ in pairs of the form $(a_i, a_{m+1-i})$, with $1 \le i \le \frac{m}{2}$. We prove that the sum of the numbers in each pair is at least $n+1$. Assuming the contrary to be true, it would exist an $i$ for which $a_i + a_{m+1-i} \le n$. But, as $i < m+1-i$, the following $i$ distinct numbers
$$a_1 + a_{m+1-i} < a_2 + a_{m+1-i} < \dots < a_i + a_{m+1-i}$$
have to belong to the set $\{a_{m+2-i}, a_{m+3-i}, \dots, a_m\}$, which has only $i-1$ elements, a contradiction. Adding
$$a_i + a_{m+1-i} \ge n+1, \quad \text{for } 1 \le i \le \frac{m}{2},$$
the conclusion follows immediately.

For $m = 2k-1, k > 2$, it can be shown, as above, that $a_i + a_{m+1-i} \ge n+1$, $1 \le i \le k-1$.
We now prove that $a_k \ge \frac{n+1}{2}$. Suppose $2a_k < n+1$. Consider
$$a_{k-1} < a_1 + a_{k-1} < a_1 + a_k < a_2 + a_k < \dots < a_{k-1} + a_k < n + 1.$$
It follows that $a_1 + a_{k-1}$, $a_1 + a_k < a_2 + a_k$, ..., $a_{k-1} + a_k$ must all belong to $A$, hence they must be equal to $a_k$, $a_{k+1}$, ..., $a_m$, respectively.
We obtain that $n+1 \le a_1+a_{2k-1} = a_1+(a_{k-1}+a_k) = (a_1+a_{k-1})+a_k = 2a_k$, which contradicts the assumption we have made.