, 2021

Solution:
In any given round, the second-best team is only eliminated if it plays against the best team. If there are $k$ teams left and the second-best team has not been eliminated, the second-best team plays the best team with probability $\frac{1}{\binom{k}{2}}$, so the second-best team survives the round with probability
$$1 - \frac{1}{\binom{k}{2}} = 1 - \frac{2}{k(k-1)} = \frac{k^2 - k - 2}{k(k-1)} = \frac{(k+1)(k-2)}{k(k-1)}.$$
So, the probability that the second-best team survives every round before the last round is
$$\prod_{k=3}^{2021} \frac{(k+1)(k-2)}{k(k-1)}$$
which telescopes to
$$\frac{\frac{2022!}{3!} \cdot \frac{2019!}{0!}}{\frac{2021!}{2!} \cdot \frac{2020!}{1!}} = \frac{2022! \cdot 2019!}{2021! \cdot 2020!} \cdot \frac{2! \cdot 1!}{3! \cdot 0!} = \frac{2022}{2020} \cdot \frac{1}{3} = \frac{337}{1010} = p$$
So,
$$\lfloor 2021 p \rfloor = \left\lfloor \frac{2021 \cdot 337}{1010} \right\rfloor = \left\lfloor 337 \cdot 2 + 337 \cdot \frac{1}{1010} \right\rfloor = 337 \cdot 2 = 674$$