Denote $A_i$ the score of the $i$-th player. Then $A_1+A_2+A_3=A_6+A_7+...+A_{13}+A_{14}$. If there is a draw, to the total score there is added $0,5+0,5=1$, otherwise there is added $0+1=1$. This means that the total score is $\binom{14}{2}$. This also applies to the total score gained by a group of players when playing each other. For example, the score gained by the first $3$ players when playing each other is $\binom{3}{2}=3$, and the score gained by the last $9$ players when playing each other is $\binom{9}{2}=36$. Then $A_6+A_7+...+A_{13}+A_{14} \geq 36$. The highest value of $A_1+A_2+A_3$ will take place if the first $3$ players win all their games with the other players. (The games beetween them don't matter because the sum of points gained will always be $3$.) Then $A_1+A_2+A_3 \leq 3+3\cdot11=36.$ We have $36 \geq A_1+A_2+A_3=A_6+A_7+...+A_{13}+A_{14} \geq 36$. This means that in order for the equality to take place, the last $9$ players lost all their games with the other players and the first $3$ players won all their games with the other players. Now we have to have find the highest numbers of draws when the two conditions from above take place. Since the first $3$ players won all their games with the other players, they can draw only with each other. If they draw all games with each other we have $3$ draws. Since $4$th and $5$th players lost against the first $3$ players and won against the last $9$ players, they can only draw between themselves. Since the last $9$ players lost all their games with the other players, they can only draw with each other. If they draw all games with each other we have $36$ draws. The answer is $3+1+36=40$.