Problem 1644

1. Define the minimum sequence: Let $m_n := \min\{a_k : k \ge n\}$ for $n = 1, 2, \ldots$. We need to show that $m_n$ is bounded.

2. Monotonicity and boundedness: Since $m_n$ is a monotone increasing sequence of integers, it must eventually become constant. Let $m$ be this constant value from some point onward.

3. **Choose a large $n$**: Take some $a_n = m$ for sufficiently large $n$. Define $M := \max\{a_{n+1}, a_{n+2}, \ldots, a_{2n}\}$.

4. Apply the given inequality: From the problem statement, we have:
$$a_n > \frac{a_{n+1} + a_{n+2} + \ldots + a_{2n}}{n + 2016}$$
Substituting $a_n = m$ and noting that $a_{n+i} \ge m$ for $i = 1, 2, \ldots, n$, we get:
$$m > \frac{(n-1)m + M}{n + 2016}$$

5. Simplify the inequality: Rearrange the inequality to solve for $M$:
$$m(n + 2016) > (n-1)m + M$$
$$mn + 2016m > nm - m + M$$
$$2016m + m > M$$
$$M < (2016 + 1)m$$
$$M < 2017m$$

6. **Contradiction if $a_k \ge m+1$**: Assume for contradiction that $a_j \ge m + 1$ for $j = n+1, n+2, \ldots, 2n$. Then:
$$m = a_n > \frac{n(m+1)}{n + 2016}$$
Simplifying this:
$$m(n + 2016) > n(m + 1)$$
$$mn + 2016m > nm + n$$
$$2016m > n$$
This is false if we choose $n > 2016m$. Hence, there must be some $k \in \{n+1, n+2, \ldots, 2n\}$ such that $a_k = m$.

7. **Boundedness of $m_n$**: To show that $m_n$ is bounded, assume for contradiction that $m_n$ is unbounded. This means for any $m$, there exists $n$ such that $a_n = m$ and $a_{n+i} > m$ for $i = 1, 2, \ldots$.

8. Minimum among larger indices: Let $m + \Delta$ be the minimum among $a_{\left\lfloor \frac{3n}{2} \right\rfloor}, \ldots, a_{2n}$. Then:
$$m = a_n > \frac{\frac{n}{2}(m+1) + \frac{n}{2}(m+\Delta)}{n + 2016}$$
Simplifying:
$$m(n + 2016) > \frac{n}{2}(m+1) + \frac{n}{2}(m+\Delta)$$
$$m(n + 2016) > \frac{n}{2}(2m + 1 + \Delta)$$
$$m(n + 2016) > n(m + \frac{1 + \Delta}{2})$$
$$2m(n + 2016) > n(2m + 1 + \Delta)$$
$$2mn + 4032m > 2mn + n + n\Delta$$
$$4032m > n + n\Delta$$
$$4032m > n(1 + \Delta)$$
$$\Delta < \frac{4032m - n}{n}$$
$$\Delta < \frac{4032m}{n} - 1$$

9. Contradiction: This implies $m + \Delta < m \left(1 + \frac{4032}{n}\right)$. Since $n$ can be made arbitrarily large, $\Delta$ can be made arbitrarily small, contradicting the assumption that $m_n$ is unbounded.

Thus, $m_n$ is bounded, and there exists a positive real $C$ such that $a_n < C$ for all positive integers $n$.

$\blacksquare$