$N$ teams take part in a league. Every team plays every other team exactly once during the league, and receives 2 points for each win, 1 point for each draw, and 0 points for each loss. At the end of the league, the sequence of total points in descending order $\mathcal{A} = (a_1 \ge a_2 \ge \cdots \ge a_N )$ is known, as well as which team obtained which score. Find the number of sequences $\mathcal{A}$ such that the outcome of all matches is uniquely determined by this information.

[I]Proposed by Dominic Yeo, United Kingdom.[/i]
1. **Define the Tournament and Score Distribution:**
   - Let $N$ teams participate in a league where each team plays every other team exactly once.
   - Each team receives 2 points for a win, 1 point for a draw, and 0 points for a loss.
   - The sequence of total points in descending order is given by $\mathcal{A} = (a_1 \ge a_2 \ge \cdots \ge a_N)$.

2. **Define a Good Tournament:**
   - A tournament $T$ is called *good* if there is no other tournament $T'$ such that $d(T) = d(T')$, where $d(T)$ denotes the final score distribution along with which teams received which score.

3. **Lemma 1: No Directed Cycles of Wins or Draws:**
   - **Proof:** If there were a directed cycle of wins or draws, we could interchange the outcomes within the cycle without changing the score distribution, which would violate the goodness of $T$. Therefore, there are no directed cycles of wins or draws. $\blacksquare$

4. **Ranking of Players:**
   - There exists a ranking $1, 2, \ldots, N$ of the players such that $i < j$ implies $i$ beats $j$ or $i$ draws $j$.

5. **Lemma 2: If $1$ Beats $2$, Then $1$ Beats Everyone:**
   - **Proof:** Suppose $1$ draws with some $x \ge 3$. If $2$ beats $x$, we can replace $\{1 \to 2, 2 \to x, 1 \sim x\}$ with $\{1 \sim 2, 2 \sim x, 1 \to x\}$, leaving the score distribution unchanged, which violates the goodness of $T$. Therefore, $1$ must beat all $x \ge 3$. $\blacksquare$

6. **Lemma 3: If $1$ Draws with $2$, Then $1$ and $2$ Beat All $x \ge 3$:**
   - **Proof:** Suppose $2$ draws with some $x \ge 3$. Then $1$ beats $x$ to avoid a cycle of draws. We can replace $\{1 \sim 2, 2 \sim x, 1 \to x\}$ with $\{1 \to 2, 2 \to x, 1 \sim x\}$, leaving the score unchanged, which violates the goodness of $T$. Therefore, $2$ must beat all $x \ge 3$. Similarly, if $1$ draws with $x \ge 3$, $2$ beats $x$, and we can replace $\{1 \sim 2, 2 \to x, 1 \sim x\}$ with $\{2 \to 1, 2 \sim x, 1 \to x\}$, which also violates the goodness of $T$. Thus, $1$ beats all $x \ge 3$. $\blacksquare$

7. **Lemma 4: Score of Top Teams:**
   - **Proof:** We have two cases:
     - **Case 1:** If $1$ beats $2$, then by Lemma 2, $1$ beats everyone, so $a_1 = 2N - 2$. Everyone else has at least one loss, so their score is at most $2N - 4$.
     - **Case 2:** If $1$ draws with $2$, then by Lemma 3, $1$ and $2$ beat everyone else, so both have a score of $2N - 3$. Everyone else has at least two losses, so their score is at most $2(N - 3) = 2N - 6$. Therefore, $a_1 = a_2 = 2N - 3$. $\blacksquare$

8. **Counting the Number of Sequences:**
   - Let $f(N)$ be the number of sequences that generate a unique tournament.
   - If $a_1 = 2N - 2$, then the person with that score beat everyone. The sequence $(a_2 \ge \cdots \ge a_N)$ must generate a unique tournament, so there are $f(N-1)$ such sequences.
   - If $a_1 = a_2 = 2N - 3$, then two players have $N-2$ wins and one draw. They must have drawn each other and beaten everyone else. The sequence $(a_3 \ge \cdots \ge a_N)$ must generate a unique tournament, so there are $f(N-2)$ such sequences.
   - Therefore, $f(N) = f(N-1) + f(N-2)$, with initial conditions $f(1) = 1$ and $f(2) = 2$.

9. **Conclusion:**
   - The number of sequences $\mathcal{A}$ such that the outcome of all matches is uniquely determined is given by the Fibonacci sequence shifted by one index: $f(N) = F_{N+1}$.

The final answer is $\boxed{F_{N+1}}$.