Can candidate who wins all pairwise contests get least score? There are $n$ voters and $m$ candidates. Each voter gives a (strict) ranking of the candidates: $a_1>a_2>\ldots>a_m$, meaning that the voter likes $a_1$ most, followed by $a_2$, etc. Then $a_1$ gets $m$ points, $a_2$ gets $m-1$ points, and so on, until $a_m$ gets $1$ point. To calculate the total points of each candidate, we just add up the points that the candidate obtains from each voter. Suppose that a certain candidate $A$ has the property that for any other candidate $X$, the number of voters who prefer $A$ to $X$ is greater than the number of voters who prefer $X$ to $A$. Is it possible that no candidate obtained a lower score than $A$?

No. Let look at the number of pairs (v,X) such as v is a voter that prefer A to X. The number of such pairs is greater than (m-1)n/2 (since for each X!=A, there exist more than n/2 of such pairs that contain X). Thus, the total number of points that A got is more than (m-1)n/2+n=(m+1)n/2 (note that if the number of such pairs that contain the voter v is k, then k gave A the score k+1. Thus, summing over all voters, the score is more than (m-1)n/2+n). The total score is m(m+1)n/2, thus the lowest score is at most (m+1)n/2, and since A got more than that, he cannot have the lowest score.