Solution:

We can assume that the initial numbers are all distinct, since this occurs with probability 1. For clarity, we denote the value of $a_{i}$ after $t$ seconds as $a_{i,t}$. The index $i$ is taken mod 2025.

In general, after $k < 1012$ seconds, we claim the expected number of distinct values remaining is $\frac{2025}{k + 1}$. To show this, we first prove that for any remaining value, its appearances are consecutive. Indeed, note that for all $i$ and $k$,

$$a_{i,k} = \max (a_{i - 1,k - 1}, a_{i,k - 1}, a_{i + 1,k - 1}) = \max (a_{i - 2,k - 2}, \ldots, a_{i + 2,k - 2}) = \dots = \max (a_{i - k,0}, \ldots, a_{i + k,0}).$$

Given an initial number $a_{c,0}$, let $j_{1}$ and $j_{2}$ be the smallest positive integers such that $a_{c - j_{1},0} > a_{c,0}$ and $a_{c + j_{2},0} > a_{c,0}$. As the initial numbers are distinct, we conclude $a_{i,k} = a_{c,0}$ if and only if $\{i - k, i - k + 1, \ldots, i + k\}$ contains $c$ but neither $c - j_{1}$ nor $c + j_{2}$ (mod 2025). The indices $i$ that satisfy this are clearly consecutive.

Now consider the indicator variables

$$\mathbf{1}_{i} = \left\{ \begin{array}{ll}1 & \mathrm{if~}a_{i,k}\neq a_{i + 1,k}\\ 0 & \mathrm{otherwise} \end{array} \right.$$

for $1\leq i\leq 2025$. Note that the number of distinct values on the board after $k$ seconds is simply $\textstyle \sum_{i = 1}^{2025}\mathbf{1}_{i}$. By linearity of expectation, it suffices to compute $\mathbb{E}[\mathbf{1}_{i}]$ for each $i$. Recall

$$a_{i,k} = \max (a_{i - k,0},\ldots,a_{i + k,0})$$

for each $1\leq i\leq 2025$. Hence, the condition $a_{i,k}\neq a_{i + 1,k}$ can be written as

$$\max (a_{i - k,0},\ldots ,a_{i + k,0})\neq \max (a_{i + 1 - k,0},\ldots ,a_{i + 1 + k,0}).$$

This is the case if and only if $\max (a_{i - k,0},\ldots ,a_{i + k + 1,0})$ is either $a_{i - k,0}$ or $a_{i + k + 1,0}$ (as $a_{i - k,0}\neq a_{i + k + 1,0}$ and $2k + 2< 2025$). Since the $2k + 2$ values $a_{i - k,0}$, ..., $a_{i + k + 1,0}$ were sampled independently from the same distribution, each of $a_{i - k,0}$ and $a_{i + k + 1,0}$ has a $\frac{1}{2k + 2}$ probability of being their maximum. Hence,

$$\mathbb{E}[\mathbf{1}_{i}] = \operatorname*{Pr}[a_{i,k}\neq a_{i + 1,k}] = \frac{2}{2k + 2} = \frac{1}{k + 1}.$$

Thus, the expected number of distinct values is

$$\sum_{i = 1}^{2025}\mathbb{E}[\mathbf{1}_{i}] = \frac{2025}{k + 1}.$$

Substituting $k = 100$ yields the answer $\left[\frac{2025}{101}\right]$