26825

I will prove by contradiction that it is possible to choose three integers from any set of five whose sum is a multiple of 3.
Firstly, split all numbers into three categories, numbers which are; multiples of 3, of the form 3x; one less than a multiple of 3, 3x-1; and one more than a multiple of 3, 3x+1.
If our five selected numbers, which we claim have no three numbers that sum to a multiple of 3, contain three numbers from the same category, then our solution is incorrect, as these three will sum to a multiple of 3. All our numbers are represented by 3x+c, where c is -1, 0 or 1, so any three numbers from the same category sum to:
3x1+c+3x2+c+3x3+c = 3(x1+x2+x3) +3c
This is clearly a multiple of 3. Therefore, we must have the 5 numbers sorted into the categories with 2, 2, 1 spread, to avoid having three numbers in the same category.
However this cannot be a solution either, as selecting one number from each category, which our numbers must contain, will also make the solution invalid, as they must sum to:
(3x1+1) + (3x2) + (3x3-1) = 3x1+3x2+3x3
This is also a multiple of three, which is a contradiction; therefore our original claim was false.
Therefore, any five integers must contain three integers whose sum is a multiple of 3.