I think, the statement somehow emphasizes on $k$, however the point (and the difficulty) lies in the second term. Suppose you have a graph and you want to make a cut (i.e. to partition its vertices into two groups) so that the cross edges (between the groups) are maximum possible. It's well known that you can make a cut so that the cross edges are at least $\frac{1}{2}|E|$. (just partition the vertices randomly, and the probability that a fixed edge is a cross edge is 1/2). Now, the question is: could we do better? It's doomed to hope that we can make the cross edges like $k|E|$ for some $k>1/2$ - just take the complete graph and see we cannot make $k$ larger than $1/2$. But still, it's possible to make a sharper estimate like $\frac{1}{2}|E|+\mathrm{ something}$. And this "something" is $|V|/4$. You cannot do better, this is sharp. Very nice and instructive problem. Anyway, All Russian, 2023, p11.10.