In a football championship with $2021$ teams, each team play with another exactly once. The score of the match(es) is three points to the winner, one point to both players if the match end in draw(tie) and zero point to the loser. The final of the tournament will be played by the two highest score teams. Brazil Football Club won the first match, and it has the advantage if in the final score it draws with any other team. Determine the least score such that Brazil Football Club has a [b]chance[/b] to play the final match.
To determine the least score such that Brazil Football Club (BFC) has a chance to play in the final match, we need to analyze the points distribution and the conditions under which BFC can qualify for the final.

1. **Total Number of Matches and Points Distribution**:
   - There are \(2021\) teams, and each team plays against every other team exactly once.
   - The total number of matches is given by the combination formula:
     \[
     \binom{2021}{2} = \frac{2021 \times 2020}{2} = 2021 \times 1010
     \]
   - Each match can result in either 3 points (win/loss) or 2 points (draw). Therefore, the total number of points distributed in the tournament is:
     \[
     3 \times 2021 \times 1010
     \]

2. **Scenario Analysis for BFC to Qualify with \(n-1\) Points**:
   - Let's denote the teams as \(T_1, T_2, \ldots, T_{2021}\), where \(T_1\) is BFC.
   - Assume \(T_{2021}\) wins all its matches against the other \(2020\) teams. Thus, \(T_{2021}\) earns:
     \[
     3 \times 2020 = 6060 \text{ points}
     \]
   - Now, consider the following match outcomes:
     - \(T_1\) (BFC) wins against \(T_2\).
     - \(T_2\) wins against \(T_3\).
     - \(T_3\) wins against \(T_4\).
     - \(\ldots\)
     - \(T_{2020}\) wins against \(T_1\).
   - All other matches end in a draw. In this scenario:
     - Each of the teams \(T_2, T_3, \ldots, T_{2020}\) earns \(3\) points from one win and \(1 \times (2020 - 2) = 2018\) points from draws, totaling:
       \[
       3 + 2018 = 2021 \text{ points}
       \]
     - \(T_1\) (BFC) earns \(3\) points from one win and \(1 \times (2020 - 2) = 2018\) points from draws, totaling:
       \[
       3 + 2018 = 2021 \text{ points}
       \]
   - Therefore, BFC can qualify for the final based on the tiebreaker rule if it has \(2021 - 1 = 2020\) points.

3. **Proof that BFC Cannot Qualify with \(n-2\) Points**:
   - Suppose BFC qualifies with at most \(2020 - 2 = 2019\) points.
   - Let \(L\) be the team with the highest score (not BFC), and let \(NL\) be any non-leader team.
   - The total number of points distributed among the \(2020\) non-leader teams is at least:
     \[
     \frac{2020 \times 2019}{2} + 1
     \]
   - By the Pigeonhole Principle, some non-leader team must earn at least:
     \[
     \frac{2020 \times 2019}{2020} + 1 = 2019 + 1 = 2020 \text{ points}
     \]
   - Since BFC must have a score greater than or equal to all non-leader teams to qualify, BFC must also have at least \(2020\) points.

Therefore, the least score such that Brazil Football Club has a chance to play in the final match is \(2020\) points.

The final answer is \(\boxed{2020}\).