<= means less or equal as there is no such symbol on my keyboard
You can prove that abcd is divisible by 3 or 5 if any of these 4 integers is divisible by 3 or 5.
Since their position is the same, let's just say that a<=b<=c<=d. Even if the order is different, for example, a<=c<=d<=b , it doesn't matter, as we just put the smallest one in front and then rename them as a, then 2nd smallest as b, and so on. It's just to simplify things.
We'll start with d being something not divisible by 3. as a, b, c, and d are positive, it's obvious that a+b+c+d > d. But since a+b+c+d is the common multiple, it must be divisible by d. a=b=c=d is not possible because if that's the case, then the smallest multiple will be d, which is less than a+b+c+d. Since a=b=c=d is not possible, a, b, or c must be less than d, then a+b+c+d < 4d while a+b+c+d is divisible by d. Then, the only possibilities left are that a+b+c+d=3d or a+b+c+d=2d.
1. If a+b+c+d=3d, while a+b+c+d is divisible by all of those four, then a, b, and c divides 3d. Is it possible that only the factors in d are scattered among a, b, and c in the same or less power such that they all divide 3d? No. Because if that's the case, then they all will divide d, and the least multiple will be d, which has been proven not possible. So the only way so that a, b, and c divides 3d while not all of them divides d is by having at least one of a, b, and c divisible by 3. So for a+b+c+d=3d, it's already proved.
2. The only remaining possibility is that a+b+c+d=2d
that way, a+b+c=d that a+b+c+d= 2(a+b+c)
in this possibility, it's obvious c<d
because a<=b<=c , a+b+c+d= 2(a+b+c)<=6c
if 2(a+b+c)=6c, then a=b=c and that means d=3a=3b=3c. This will cause d being the least common multiple again. So this possibility doesn't even fulfill the requirement.
Then, 2(a+b+c)<6c. And it's obvious that 2(a+b+c)>2c . Since a+b+c+d = 2(a+b+c) is divisible by c, the remaining possibilities are
2(a+b+c) = 3c, 4c, or 5c
don't forget that in this second possibility a+b+c=d, then if 2(a+b+c)=3c or 5c, it will cause d to be divisible by 3 or 5 as well.
Now the remaining unproven possibility is 2(a+b+c)=4c that means a+b = c
the total sum now becomes a+b+c+d= 2(a+b+ (a+b)) = 4(a+b)
we now know that b divides a+b+c+d= 4(a+b)=4a+4b , which means that b divides 4a
as b is more or equal a, while 4a/a=4 , 4a/b <=4a/a=4
that means 4a/b=4 or 4a/b=3 or 4a/b=2 or 4a=b
For 4a/b=4 , a=b , which cause c=2a . That way, a and b divides c, and c divides d. That way, all of them divides d, and the least common multiple becomes d again, and it fails the requirement.
For 4a/b=3 , we get 3 divides 4a, which means 3 divides a.
For 4a/b = 2, b=2a, and then c= a+b= 3a. That means 3 divides c.
For 4a/b=1, b=4a , and c=a+b=5a. That means 5 divides c.
With this, we find out that all the options that can possibly fulfill the requirement of a+b+c+d being the least common multiple of a, b, c, and d lead to one of a,b,c, or d being divisible by 3 or 5. That way, abcd will definitely be divisible by 3 or 5.