Problem 4. A set of at least three positive integers is called uniform if removing any of its elements the remaining set can be disjoint into two subsets with equal sums of elements. Find the minimal cardinality of a uniform set.

Peter Boyvalenkov, Emil Kolev
4. Let $A=\left\{a_{1}, a_{2}, \ldots, a_{n}\right\}$ be a uniform set. Set $S=a_{1}+a_{2}+\cdots+a_{n}$. It follows from the given condition that $S-a_{i}$ is an even number for any $i=1,2, \ldots, n$ Suppose that the number $S$ is even. Then all the numbers $a_{i}$ are even. Set $a_{i}=2 b_{i}, i=1,2, \ldots, n$. Then it is easy to see that the set $B=\left\{b_{1}, b_{2}, \ldots, b_{n}\right\}$ is uniform, too. So, we may assume that $S$ is an odd number and whence $a_{1}, a_{2}, \ldots, a_{n}$ and $n$ are also odd numbers.

We shall prove that $n=7$. It is not difficult to check that $\{1,3,5,7,9,11,13\}$ is a uniform set. It remains to show that there are no unform sets with 5 elements, since it is is obvious that the sets with 3 elements are not uniform.

Suppose that $A=\left\{a_{1}, a_{2}, a_{3}, a_{4}, a_{5}\right\}$ is a uniform set and let $a_{1}<a_{2}<$ $a_{3}<a_{4}<a_{5}$. Considering the set $A \backslash\left\{a_{1}\right\}$ we see that either $a_{2}+a_{5}=a_{3}+a_{4}$ or $a_{2}+a_{3}+a_{4}=a_{5}$. Considering the set $A \backslash\left\{a_{2}\right\}$ we get that either $a_{1}+a_{5}=a_{3}+a_{4}$ or $a_{1}+a_{3}+a_{4}=a_{5}$.

- If $a_{2}+a_{5}=a_{3}+a_{4}$ and $a_{1}+a_{5}=a_{3}+a_{4}$, then $a_{1}=a_{2}$.
- If $a_{2}+a_{5}=a_{3}+a_{4}$ and $a_{1}+a_{3}+a_{4}=a_{5}$, then $a_{1}=-a_{2}$
- If $a_{2}+a_{3}+a_{4}=a_{5}$ and $a_{1}+a_{5}=a_{3}+a_{4}$, then $a_{1}=-a_{2}$.
- If $a_{2}+a_{3}+a_{4}=a_{5}$ and $a_{1}+a_{3}+a_{4}=a_{5}$, then $a_{1}=a_{2}$.

Since all the above possibilities lead to a contradiction, we conclude that there are no unform sets with 5 elements.