Strange Pigeons and Stranger Holes

Obviously, you can't pick a multiple of 3, so your numbers must leave a remainder of 1 or 2 when divided by 3. But if you had a remainder of 1 AND a remainder of 2, you could just add those two numbers and have a multiple of 3.

Thus, your numbers must either all leave remainders of 1, or all leave remainders of 2. However, if you take any 3 of these, they will sum to a multiple of 3, since $3\cdot 1 = 3$ and $3\cdot 2=6.$ Hence, it's impossible!

---

BONUS: In fact, if you pick any $n$ whole numbers, there is always a subset that sums to a multiple of $n-1.$

Call the numbers $A_1, A_2, \ldots,A_n.$

Then, consider the sums:
$B_1 = A_1$

$B_2 = A_1 + A_2$

$\ldots$

$B_n= A_1 + A_2 + \ldots+ A_n$

Note that, since there are only $n-1$ possible remainders when dividing by $n-1$ , at least two of these sums must leave the same remainder (this is a result of the Pigeonhole Principle ).

Thus, the difference between those two sums will be divisible by $n-1$ , but the difference between those sums is a sum of a group of our $A_i$ ! For example, if $B_4$ and $B_6$ leave the same remainder, then $B_6 - B_4 = A_5+A_6$ is divisible by $n-1$ .

A good solution has been provided by Eli, so here are random comments instead:

Since group is not specified, I just assume it's subset . This means we don't even need any numbers, since an empty subset already sums to a multiple of 3.

For this particular case, it's actually sufficient to take 3 numbers instead of 4; it's guaranteed that some non-empty subset sums up to a multiple of 3.