The question is related to the famous locker puzzle:
The director of a prison offers 100 prisoners on death row, which are numbered from 1 to 100, a last chance. In a room there is a cupboard with 100 drawers. The director puts in each drawer the number of exactly one prisoner in random order and closes the drawers afterwards. The prisoners enter the room one after another. Each prisoner may open and look into 50 drawers in any order and the drawers are closed again afterwards. If during this search every prisoner finds his number in one of the drawers, all prisoners are pardoned. If just one prisoner does not find his number, all prisoners have to die. Before the first prisoner enters the room, the prisoners may discuss their strategy, afterwards no communication of any means is possible. What is the best strategy for the prisoners?
I am interested in the modification where the prisoners are only allowed to use a predetermined strategy: each of them must choose the 50 boxes he will open before entering the room.
In this case I believe the optimal strategy is the following one. Divide the prisoners into two groups, say, with numbers 1-50 and 51-100. Each prisoner from the first group should open boxes 1-50, from the second, 51-100.
More generally, if each of $nm$ prisoners is allowed to open $m$ boxes, then the best strategy seems to be the same: divide into groups $1$ to $m$, $m+1$ to $2m$, ... $(n-1)m+1$ to $nm$ and let each prisoner open the boxes with numbers from his group.
This is easily seen to be optimal for $m=2$. Is this so for any $m$? Is there any reference?