Problem 1220

I. solution. If 0 is among the numbers, it is enough to find two numbers with equal absolute values next to it. Since the absolute value of the other numbers can only be 99 different, and there are 100 numbers different from zero, according to the pigeonhole principle, we can indeed find such a pair of numbers. Thus, in the following, we assume that the numbers are different from zero.

Since the positive numbers and the negatives can only be 99 different, among the 101 numbers, there must be both positive and negative. Let's denote the smallest number by $A$ and the largest by $B$:

$$A<0<B$$

If we add the positive numbers to $A$ one by one, we get integers larger than $A$, the largest of which is $A+B$. Moving on, now let's add the negative numbers to $B$ in sequence, thus getting integers larger than $(A+B)$ and smaller than $B$. In total, we have 100 different sums, since every number different from $A$ and $B$ appears in one sum, and the pair $(A, B)$ also gives us one sum. Let's denote the set of numbers by $H$, the set of sums we just created by $S$, and the set of the original numbers multiplied by $(-1)$ by $G$.

In $H$ and $G$, there are 101 different integers with an absolute value less than 100, and in $S$ there are 100, also with an absolute value less than 100. Since the number of integers with an absolute value less than 100 is 199, $G$ and $S$ must have at least two common elements. Let's denote one of these by $F$, and let the sum in $S$ equal to $F$ be $C+D$, where $C$ is one of the numbers $A, B$, and $D$ is a number with a different sign from $C$:

$$F=C+D$$

Since by definition of $G$, $(-F)$ is one of the original numbers, we are done if $(-F)$ is different from both $C$ and $D$. If $(-F)$ is equal to one of $C, D$, then the other of $C, D$ is equal to $(-F)$ doubled, which is certainly the one with the larger absolute value of $C$ and $D$. If this larger absolute value number is part of the sum, then it is certainly one of $A, B$ (which does not necessarily exclude the other from being one of them). Since $G$ and $S$ have at least two different common elements, our reasoning so far only fails to lead to the desired zero-sum triplet if both $A$ and $B$ are even, and both $A / 2$ and $B / 2$ are among the original numbers. But then the absolute value of the original numbers is less than 99, and the same is true for the elements of $G$ and $S$. Therefore, in this case, $G$ and $S$ have more than two different elements, and among these, there is certainly one that leads to a zero-sum triplet with different elements.

II. solution. Let's imagine a hundred small urns and label them with the integers from 0 to 99. Let's put each of the given numbers into the urn labeled with its absolute value. If the urn labeled 0 is not empty, let's find an urn next to it that has two numbers, and we are done. Such an urn is certainly there because the number of urns is less than the number of numbers (and of course, at most one number can go into the urn labeled 0.)

If the urn labeled 0 is empty, let's set it aside and find the one with the largest label among those that have two numbers. Since now the number of numbers is two more than the number of urns, there are still more numbers than urns in the remaining urns. Let's put the selected urn in front of us, place the urns with larger labels to the right, and the smaller ones to the left.

There is at most one number in the urns on the right, so the number of urns on the left is less than the number of numbers in them. So if we now select the positive and negative numbers separately from the urns on the left, we will find at least $B / 2$ more numbers in at least one of the sets, where $B$ is the number of urns on the left. Now let's pair the urns on the left so that the sum of the labels of the paired urns is $(B+1)$. If $B$ is even, we are done, because then we will certainly find a pair in which there are numbers with the same sign, say $C$ and $D$, and the absolute value of their sum is $B+1$. Since the urn labeled $(B+1)$ is in front of us and contains two numbers, we are done if we take a number with a different sign from $C$ and $D$ from this urn. We are also done if $B$ is odd, but we still find a pair of urns in which there are numbers with the same sign. So we only need to examine the case where $B$ is odd, and in the paired urns, there are only two numbers with different signs. There are certainly this many, because even in this case, there can be at most $(B-1)$ numbers in the paired urns on the left, and if the unpaired urn labeled $(B+1) / 2$ also contains the maximum possible, that is, two numbers with different signs, there are still only $(B+1)$ numbers in the urns on the left. This can only happen if there is exactly one number in each of the urns on the right, so there is also a number in the urn labeled $(B+2)$. For the sake of easier discussion, let's assume this number is negative. We already know that in our case, there are exactly $A=(B+1) / 2$ positive and negative numbers in the urns on the left. Now let's pair the urns whose label sum is $(B+2)$, and include the urn in front of us among the urns on the left. This way, we get $A$ pairs, in which there are $(A+1)$ positive numbers. So there is a pair with two positive numbers, and by taking the negative number from the urn labeled $(B+2)$, we get a suitable triplet of numbers.

Remark. Both solutions can be reformulated for the general case where there are $(2 n+1)$ different numbers with an absolute value less than $2 n$. The fact that parity plays a significant role here is well illustrated by the annoying conclusions of the solutions, that when we are about to reach our goal, we still have to deal with some exceptional cases. As the example $\{ \pm n, \pm(n+1), \ldots, \pm 2 n\}$ shows, $(2 n+2)$ different numbers with an absolute value less than $(2 n+1)$ can already be given so that no three of them have a sum of zero.