Let $n$ be the number of players. Considering the total number of points, we get $$\binom{n}{2}\geqslant k+\left( k+\frac{1}{2}\right)+\cdots +\left( k+\frac{n-1}{2}\right)\implies n\geqslant 4k+1.$$Let the highest score be $m$. We get $$\binom{n}{2}\leqslant m+\left( m-\frac{1}{2}\right)+\cdots +\left( m-\frac{n-1}{2}\right)\implies m\geqslant \frac{3}{4}(n-1).$$Let $w,d,\ell$ be the number of wins, draws, and loses of the player with the highest score. $$m=w+\frac{1}{2}d\leqslant \frac{1}{2}w+ \frac{1}{2}(n-1)\implies w\geqslant \frac{n-1}{2}\geqslant 2k.$$ For equality case, consider players $P_1,P_2,\dotsc, P_{4k+1}$. For $i<j$, $P_i$ wins against $P_j$ if $j\geqslant i+(2k+1)$, otherwise the match is a draw.