USAMO

For $k \ge 1$, let
$$s_k = a_1 + a_2 + \cdots + a_k.$$
We have
$$\frac{s_{k+1}}{k+1} < \frac{s_{k+1}}{k} = \frac{s_k + a_{k+1}}{k} \le \frac{s_k + k}{k} = \frac{s_k}{k} + 1.$$
On the other hand, for each $k$, $s_k/k$ is a positive integer. Therefore
$$\frac{s_{k+1}}{k+1} \le \frac{s_k}{k},$$
and the sequence of quotients $s_k/k$ is eventually constant. If $s_{k+1}/(k+1) = s_k/k$, then
$$a_{k+1} = s_{k+1} - s_k = \frac{(k+1)s_k}{k} - s_k = \frac{s_k}{k},$$
showing that the sequence $a_k$ is eventually constant as well.

For $k \ge 1$, let
$$s_k = a_1 + a_2 + \cdots + a_k \quad \text{and} \quad \frac{s_k}{k} = q_k.$$
Since $a_k \le k - 1$, for $k \ge 2$, we have
$$s_k = a_1 + a_2 + a_3 + \cdots + a_k \le n + 1 + 2 + \cdots + (k-1) = n + \frac{k(k-1)}{2}.$$
Let $m$ be a positive integer such that $n \le \frac{m(m+1)}{2}$ (such an integer clearly exists). Then
$$q_m = \frac{s_m}{m} \le \frac{n}{m} + \frac{m-1}{2} \le \frac{m+1}{2} + \frac{m-1}{2} = m.$$
We claim that
$$q_m = a_{m+1} = a_{m+2} = a_{m+3} = a_{m+4} = \dots$$
This follows from the fact that the sequence $a_1, a_2, a_3, \dots$ is uniquely determined and choosing $a_{m+i} = q_m$, for $i \ge 1$, satisfies the range condition
$$0 \le a_{m+i} = q_m \le m \le m+i-1,$$
and yields
$$s_{m+i} = s_m + iq_m = mq_m + iq_m = (m+i)q_m.$$

For $k \ge 1$, let
$$s_k = a_1 + a_2 + \cdots + a_k.$$
We claim that for some $m$ we have $s_m = m(m-1)$. To this end, consider the sequence which computes the differences between $s_k$ and $k(k-1)$, i.e., whose $k$-th term is $s_k - k(k-1)$. Note that the first term of this sequence is positive (it is equal to $n$) and that its terms are strictly decreasing since
$$(s_k - k(k-1)) - (s_{k+1} - (k+1)k) = 2k - a_{k+1} \ge 2k - k = k \ge 1.$$
Further, a negative term cannot immediately follow a positive term. Suppose otherwise, namely that $s_k > k(k-1)$ and $s_{k+1} < (k+1)k$. Since $s_k$ and $s_{k+1}$ are divisible by $k$ and $k+1$, respectively, we can tighten the above inequalities to $s_k \ge k^2$ and $s_{k+1} \le (k+1)(k-1) = k^2 - 1$. But this would imply that $s_k > s_{k+1}$, a contradiction. We conclude that the sequence of differences must eventually include a term equal to zero.
Let $m$ be a positive integer such that $s_m = m(m-1)$. We claim that
$$m - 1 = a_{m+1} = a_{m+2} = a_{m+3} = a_{m+4} = \dots$$
This follows from the fact that the sequence $a_1, a_2, a_3, \dots$ is uniquely determined and choosing $a_{m+i} = m-1$, for $i \ge 1$, satisfies the range condition
$$0 \le a_{m+i} = m-1 \le m+i-1,$$
and yields
$$s_{m+i} = s_m + i(m-1) = m(m-1) + i(m-1) = (m+i)(m-1).$$