If you google "Knight distance", there will come a few pages with a general formula that's useful for this problem.

Let f(x, y) be the function that computes the distance in knight moves from point (0, 0) to point (x, y) with x ≥ y (if not, just swap them, the cartesian plane is symmetrical), and let d = x - y. Then the formula is...

(when y > d)

(when y ≤ d)

Once you know that, you can solve it with min-cost max-flow. Build a bipartite graph. Left nodes are knights and right nodes are points. Join each left node with each right node by setting capacity to 1 and cost equal to that knight distance to that point. Also join the source to every left node and all right nodes to the sink with capacity 1 and cost 0.