2023 AMC 12A Problem 13

Let there be $L$ left-handed and $2L$ right-handed players, for $3L$ players and $\binom{3L}{2}$ games total. If right-handers win $R$ games, left-handers win $1.4R,$ so the total is $2.4R=\tfrac{12}{5}R.$ For this to be an integer count, the total number of games must be a multiple of $12.$ The left-handers can win at most every game involving at least one left-hander, namely $\binom{L}{2}+2L^2=\tfrac{5L^2-L}{2}.$ On the other hand, they must win $\tfrac{7}{12}\binom{3L}{2}=\tfrac{7L(3L-1)}8$ games. Comparing these quantities gives $21L-7\le 20L-4,$ so $L\le 3.$ Testing $L=1,2,3$ gives totals $3,15,36;$ only $36$ is a multiple of $12.$ It is achievable: the $3$ left-handers can win all $18$ mixed games and their $3$ internal games, giving $21=1.4\cdot 15$ wins. Thus, the correct answer is B.