Consider all the possible divisions, and in each case count up the total number of cases E in which a member has an enemy in his own house. The division in which E is minimized must have the required property. If it didn’t, that would mean that some member has at least two enemies in his own house. In that case he’d have one enemy at most in the other house, and could reduce E further by moving to the opposite house. That’s a contradiction, so the required task must be possible.