Take the standard graph theory interpretation, where students and vertices and two vertices are connected IFF the students are friends. And let a garden on n vertices be a full graph of n vertices. We say 2 gardens are not connected if they don’t share a common vertex, and there is no edge between the gardens. A graph is good if the final condition is fulfilled. I claim that n = 1011. Lower bound: Notice that if our graph is an union two not connected gardens who’s size adds up to 2022, then this condition will remain (invariant). Hence if our starting graph is an union of two 1011 gardens, it will take at least 1011 moves to remove one of them. Upper bound: If every vertex has degree >1010 we are done, because the graph is good. Now assume otherwise. Take a vertex with degree x<=1010. Use the first x moves to make the degree of the vertex 0, and use another move to make degree of the vertex 2021. Notice that this graph is good and we have used at most x+1<=1011 moves.