1. Define the Problem and Variables:
- There are 14 players, each playing against every other player.
- Points: Win = 1, Loss = 0, Tie = $\frac{1}{2}$.
- The best three players (set $A$) have the same total points as the last nine players (set $C$).
- The number of ties is maximal.

2. Total Number of Games:
- Each player plays against 13 others.
- Total number of games:
$$\binom{14}{2} = \frac{14 \times 13}{2} = 91$$

3. Points Distribution:
- Total points in the tournament:
$$91 \text{ games} \times 1 \text{ point per game} = 91 \text{ points}$$
- Let $f(X)$ be the total points of players in set $X$.
- Given: $f(A) = f(C)$.

4. **Points Calculation for Sets $A$ and $C$:**
- Points for set $A$:
$$f(A) \leq 3 + 3 \times 11 = 36$$
(3 games among themselves and 11 games each against others)
- Points for set $C$:
$$f(C) \geq 36$$
(36 games among themselves)

5. Equality Condition:
- Since $f(A) = f(C)$, we have:
$$f(A) = 36 \quad \text{and} \quad f(C) = 36$$
- Players in $A$ win all games against $B \cup C$.
- Players in $B$ win all games against $C$.

6. Games within Sets:
- **Set $A$:**
- Cannot have all three players draw (would result in same score).
- Possible scenario: $v_1 \to v_3$, $v_1 \leftrightarrow v_2$, $v_2 \leftrightarrow v_3$.
- Result: $v_1 > v_2 > v_3$ with 2 draws.

- **Set $C$:**
- Net games won by nine players must sum to zero.
- At least 4 distinct $w_i$ with the same sign.
- Minimum sum of positive $w_i$: $1 + 2 + 3 + 4 = 10$.
- Possible scenario: $v_9 \to \{v_1, v_2, v_3, v_4\}$, $v_8 \to \{v_1, v_2, v_3\}$, $v_7 \to \{v_1, v_2\}$, $v_6 \to v_1$.
- Result: 10 games won, 26 draws.

7. **Set $B$:
- Unique pair with the same score.
- Additional draw: 1 draw.

8. Total Number of Draws:**
- Draws in set $A$: 2
- Draws in set $C$: 26
- Draws in set $B$: 1
- Total draws:
$$2 + 26 + 1 = 29$$

The final answer is $\boxed{29}$