Iranian Mathematical Olympiad

First, we prove that the sequence is bounded; divide the sequence into blocks of $2025$ terms. It can be easily seen that the maximum of these blocks is decreasing. If the infimum of the maximums of the blocks is zero, the statement follows. Therefore, assume this infimum is $M$. Due to the decreasing nature of the maximums, it follows that the infimum of the maximums of all consecutive blocks of $2025$ terms (not just the initial blocks) is equal to $M$. In fact, beyond a certain point, the maximum of each $2025$-term block will be between $M$ and $M + \epsilon$.
This implies that the minimum in each of these blocks is smaller than $\epsilon$. Consider a term in the sequence from this point onwards that is smaller than $\epsilon$. According to the inequality above, each of the next $2025$ terms will be in the interval $[M - \epsilon, M + \epsilon]$.
Suppose the largest of these terms is $M + \epsilon$ and the smallest of these terms is $M + \epsilon'$. If $\epsilon' < 0$, then the $2026$th term will be greater than $\epsilon$, which implies that the supremum of the next $2025$ terms becomes less than $M$, which is a contradiction.
Therefore, from a certain point onwards, the sequence will consist of $2025$ terms larger than $M$ and one small term. More precisely, the sequence will be of the form
$$M + \epsilon_1, M + \epsilon_2, \dots, M + \epsilon_{2025}, \epsilon = \max(\epsilon_i) - \min(\epsilon_i)$$
In the next block, the first term will be larger than all terms. Therefore, we can assume $\epsilon_1 > \epsilon_i$. This easily implies that if the next block is
$$M + \epsilon'_1, M + \epsilon'_2, \dots, M + \epsilon'_{2025}, \epsilon' = \max(\epsilon'_i) - \min(\epsilon'_i)$$
we have
$$\epsilon'_1 = \epsilon_1 - \epsilon, \epsilon'_i < \epsilon_1 - 2\epsilon \implies \epsilon' \ge \epsilon$$