Yes all four numbers must be odd.
Dr. Peterson asked how you know that the units digit in a certain power of 2 is 4. Here is a way to prove it using weak mathematical induction, a technique you probably do not know if you are in primary school.
What we do is that we first prove a proposition is true for a single integer (often 1). We then know that the proposition is true for at least one integer. Take any integer for which the proposition is true (I usually call it k). Now prove it is true for k + 1. If so, it is true for the first number we showed, then the next, then the next, and so on forever and ever.
[MATH]2^1 = 2, \ 2^2 = 4,\ 2^3 = 8,\ 2^4 = 16, \ 2^5 = 32. [/MATH]
So we know that
[MATH]n = 1 \implies 2^{4n - 2} = 2^2 = 10 * 0 + 4 = 10i_n + 4 \text { and } i_n \in \mathbb N_{\ge 0}.[/MATH]
[MATH]k \text { is any integer} \ge 1 \text { and such that }[/MATH]
[MATH]2^{(4k-2)} = 10i_k + 4 \text { and } i_k \in \mathbb N_{\ge 0}.[/MATH]
[MATH]2^{\{4(k+1)- 2\}} = 2^{(4k+4-2)} = 2^{\{(4k-2)+4\}} =[/MATH]
[MATH]2^{(4k-2)} * 2^4 = (10i_k + 4) * 16 = 16(10i_k) + 64 = [/MATH]
[MATH]10(16i_k) + 10(6) + 4 = 10(6 + 16i_k) + 4.[/MATH]
[MATH]i_k \text { is a non-negative integer} \implies 6 + 16i_k \text { is too.}[/MATH]
[MATH]\text {Let } i_{k+1} = 6 + 16i_k \implies[/MATH]
[MATH]2^{\{4(k+1)-2)} = 10i_{(k+1)} + 4 \text { and } i_{(k+1)} \in \mathbb N_{\ge 0}.[/MATH]
[MATH]\therefore n \in \mathbb N_{> 0} \implies 2^{4n - 2} = 10i_n + 4 \text { and } i_n \in \mathbb N_{\ge 0}.[/MATH]
All that remains to do is to show that
[MATH]1002 = 1004 - 2 = 4 * 251 - 2.[/MATH]
And clearly 251 is a non-negative integer.
I doubt a primary school student is expected to do that proof. (Or you could use modular arithmetic if you know that).
Can all four numbers equal 1?
If just three of the numbers are 1, what is the unit digit of the sum of all four numbers?