What if in Guess another's present problem introduce more present's colours? Also i would reduce a number of men to number of colours. So the task basically the following:
N clever men receive presents from the president. There are N possible colors of present. Each man knows the color of his own gift only.
Then each man must guess a colour of a gift of some man: he must chose a man (besides himself) and a colour, write these down on a piece of paper and give this paper to an organiser. Once all this is done the organiser counts amount of correct guesses out of N.
The men know all the described procedure in advance and have time to develop a strategy, before receiving any presents. Once they receive the presents, they will be unable to communicate with each other.
Their task is to guarantee maximum amount of correct guesses. Your task is to say 1) what is this maximum amount, 2) what can be a strategy of the men and 3) prove that there is no other strategy, which can guaranty a bigger amount.
@Mike Earnest gave a quite clear proof that for such a task it is impossible to guarantee more than N/N=1 correct guess. When N=2 it is quite easy to guarantee 1 correct guess: see answer of @The Dark Truth. But already with N=3 i'm failing to figure out a strategy.
Also note that the task is quite similar to well known N logicians wearing hats of N colors problem. The differences are: 1. a man sees only his "hat", not "hats" of all others, 2. a man must guess another person color and he can chose this person, he is not stuck with guessing his own "hat" color.