Answer: $20$.

We will estimate the number $S$ — the sum of the numbers of draws for all $8$ chess players. This sum is exactly twice the number of draws in the tournament (since each draw is counted twice — for both players).

We will prove that $S \le 41$ — then the number of drawn games in the tournament does not exceed $20$, since it is an integer. This number is indeed the answer, since an example with $20$ draws exists (see Fig. 8; the players are denoted by the letters $A$, $B$, $C$, $D$, $E$, $F$, $G$, and $H$).

Note immediately that if two people did not lose to anyone, then they played a draw with each other and, according to the condition, have different numbers of points; the same is true if they did not win against anyone. It follows that there is at most one person with $7$ draws and at most $2$ people with $6$ draws (such a person either did not lose to anyone or did not win against anyone). In addition, there are at most one person with $5$ draws, all of whose two decisive games were both won or both lost. Note that all other players with $5$ draws have $3\frac{1}{2}$ points.

Suppose there is a person $X$ with $7$ draws; then there cannot be other players with $5$ draws, because they would have the same number of points as $X$, and they would have played a draw with $X$. Therefore, in this case, there are at most $2$ people with $5$ draws, and $S \le 7 + 2 \cdot 6 + 2 \cdot 5 + 3 \cdot 4 = 41$.

Now suppose there is no person with $7$ draws. Estimate in this case the number of players with $5$ draws and $3\frac{1}{2}$ points. Each of them must have played with each other not in a draw; however, they have only $2$ decisive games — so there are at most $3$ of them, and in total, the number of players with $5$ draws does not exceed $3 + 2 = 5$. Therefore, in this case, $S \le 2 \cdot 6 + 5 \cdot 5 + 4 = 41$, as required.