Turn the problem around: imagine that the $n$ people are sitting in a row, and we’re to distribute $k-n$ seats in the row in such a way that there is at least one seat in each gap between two adjacent people. There are $n-1$ gaps between the people, and we can also put seats on the ends, so we’re distributing $k-n$ seats in $n+2$ slots with the restriction that the middle $n-1$ slots must each get at least one seat. Of course this means that $k-n$ must be at least $n-1$, so $k\ge 2n-1$.
Start by putting one seat in each of the $n-1$ gaps between people; that leaves us with $$k-n-(n-1)=k-2n+1$$ seats that can be freely distributed amongst the $n+1$ slots. This is a standard stars-and-bars problem, whose solution is