For the opponent computers of $a_i$, we look at the $a_j$ they are playing against, see the following table.

$b_i:$
| $a_{i-4}$ | $a_{i-3}$ | $a_{i-2}$ | $a_i$ |
|---|---|---|---|

$b_{i+2}:$
| $a_{i-2}$ | $a_{i-1}$ | $a_i$ | $a_{i+2}$ |
|---|---|---|---|

$b_{i+3}:$
| $a_{i-1}$ | $a_i$ | $a_{i+1}$ | $a_{i+3}$ |
|---|---|---|---|

$b_{i+4}:$
| $a_i$ | $a_{i+1}$ | $a_{i+2}$ | $a_{i+4}$ |
|---|---|---|---|

Note that computers $a_{i-2}$, $a_{i-1}$, $a_{i+1}$ and $a_{i+2}$ each have two common opponent computers with $a_i$. Furthermore, $a_{i-4}$, $a_{i-3}$, $a_{i+3}$ and $a_{i+4}$ each have one common adversary computer with $a_i$. Only for the first two, $a_{i-4}$ and $a_{i-3}$, the common opponent with $a_i$ is the same, namely $b_i$.

Now suppose one player chose the same computer for the second round, say the player on $a_{2025}$. We will now prove by induction that all players on $a_i$ and $b_i$ have chosen the same computer for the second round.

Per the induction hypothesis, suppose that the player on a certain computer $a_i$ is the same in both rounds. Since everyone has the same opponents in both rounds, the players who were on $a_{i-4}$ and $a_{i-3}$ again each have the same common co-player in the game with $a_i$, namely the player who was on $b_i$. This means that the player on $b_i$ also chose the same computer. Now we note that only the player who was on $a_{i-2}$ has a second common opponent with $a_i$ in addition to $b_i$. So also on $a_{i-2}$, the same player chose the same computer again.

Since 2025 is odd, it now follows with induction that everyone has played on the same computer, as soon as any one player has played on the same computer. $\square$