There do not exist a set of five whole numbers that doesn't include three that add up to a multiple of three.
Proof:
For every integer (whole number, including positive numbers, 0, and negative numbers), it can be expressed as 3n-1, 3n or 3n+1 (n is an integer).
It is because when you divided a number by 3, it can have remainder 0, 1 or 2. For remainder 0, that number is divisible by 3, i.e. 3n.
For remainder 1, that number is larger than a multiple of three by 1, i.e. 3n+1.
For remainder 2, that number is larger than a multiple of three by 2, which is equivalent to smaller than a multiple of three by 1, i.e. 3(n-1)+2 = 3n-1.
Denote numbers which can be expressed as 3n-1, 3n and 3n+1 by [-1], [0] and [1] respectively.
For example, 89 = 3*29+2 = 3*30-1, so 89 can be represented by [-1]
The sum of three numbers we choose is divisible by 3 if and only if the sum of the numbers in the [ ] bracket, which I defined earlier, is divisible by 3.
For example, since 8 is [-1], 12 is [0], 13 is [1], (-1)+0+1=0, which is divisible by three, so 8+12+13=33=3*11 is also divisible by 3.
When we choose 5 integers, there are only 3 cases.
Case 1: All three types are present, i.e. [-1], [0], [1] are present.
Then we can just choose these three types and add them together. Since the sum of the numbers in the [ ] bracket is 0, which is divisible by 3, the sum of those numbers are also divisible by 3.
Case 2: Exactly two types are present.
When there is only one number of a type, then there are four numbers of another type.
When there are two number of a type, then there are three numbers of another type. Therefore, three of more numbers must be of the same type.
Choose the three numbers of the same type, the sum of the numbers in the [ ] bracket must be divisible by 3, since the number in the [ ] bracket is the same and appeared three times. Therefore, the sum of those numbers is divisible by 3.
Case 3: Only one type is present, i.e. all numbers are of the same type.
Just choose any three of them. The sum of the numbers in the [ ] bracket must be divisible by 3, since the number in the [ ] bracket is the same and appeared three times. Therefore, the sum of those numbers is divisible by 3.
Combining all these cases, there do not exist a set of five whole numbers that doesn't include three that add up to a multiple of three.