Yes, it's NP-hard via this reduction from CLIQUE.
Let (G, k) be the instance of CLIQUE (determine whether there exists a clique of cardinality at least k). Prepare a k-partite graph H with k copies of each vertex in G and edges between two vertices v and w if and only if they are in different parts and they are copies of adjacent vertices in G. There exists a k-clique in G if and only if there exists a k-clique in H. (With weights: make the existing edges weight 1 and introduce the missing ones with weight 0.)
(Surely this is in the literature, but it's Sunday, and I don't feel like looking.)