A box contains several slips of paper, each with a natural number written on it. We know that no matter how we pick three slips, there will be two among them such that the sum of the numbers on them is divisible by 5. What is the maximum number of slips that can have a number not divisible by 5?
Let's separate the numbers in the box that are not divisible by 5 according to the remainder they give when divided by 5. We can have the same remainder on at most two cards, because if any three of them give the same remainder \( m \) when divided by 5, then the sum of any two of them will give the same remainder as \( 2m \) when divided by 5. However, if \( m \) is not divisible by 5, \( 2m \) cannot be divisible by 5 either. This also implies that if there are two cards giving the same remainder, we can only choose a third card that complements them to a multiple of 5. Therefore, either all four possible remainders can occur only once, or only two different remainders occur, each at most twice. In both cases, there can be at most four cards with numbers not divisible by 5. Both cases can indeed occur, as shown by the examples \(\{1,2,3,4\},\{1,1,4,4\}\).