Anyone got solution to this?

Lemma: For any $4k$ points on a circle, with $2k$ coloured red and $2k$ coloured blue, there exists a way to draw $2k$ chords joining distinct coloured points such that no $2$ intersect. Proof: Consider a drawing that minimizes the total sum of lengths of chords. Suppose two of them intersect $AB, CD$, where $A,C$ red and $B,D$ blue. Then change it to $AD, BC$, which clearly decreases the sum of lengths. Thus this drawing cannot have any intersection. Apply the lemma by colouring $\{1,...,k,3k+1,...,4k\}$. red and the rest blue. Then the max difference is $3k-1$. For part ii, write $k+2,k+3,...,3k-2,3k-1,k+1,4k,k,4k-1,...,3k+1,1,3k$ in clockwise order. This means that some two numbers in $k+1,4k,...,1,3k$ are connected, and they would come from each of $1,...,k+1$ and $3k,..,4k$. Not hard to show that this will cause two numbers differing by $3k-1$ to be connected by a chord.