I believe the follow process of division of $ ABCD $ into $ n$ ($ {n}\equiv1\bmod 3 $) smaller squares, will give, for each $ n $, the maximum sum of the perimeters of smaller squares that intersect $ AC $: 1) Divide $ ABCD $ into 4 equal squares 2) Divide the smaller square which has $ A $ as vertex into 4 equal squares 3) Divide the smaller square which has $ C $ as vertex into 4 equal squares 4) etc In each process we will divide one of the smaller squares, whose main diagonal belongs to $ AC $ into 4 equal squares. Actually this division will be done in the direction $ A $ to $ C $ and after dividing the smaller square which has $ C $ as one vertex, we will start the process again, dividing the smaller square which has $ A $ as a vertex into 4 equal squares. Note that in this way all squares intersect $ AC $ in at least one point. Does anyone know if the process above gives the maximum sum and how to prove this? In the positive case the bound 1500 can be improved.