Solution:
Let $A=\{a_{1}, a_{2}, \ldots, a_{n}\}$ 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 $B=\{b_{1}, b_{2}, \ldots, b_{n}\}$ 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 uniform sets with 5 elements, since it is obvious that the sets with 3 elements are not uniform.

Suppose that $A=\{a_{1}, a_{2}, a_{3}, a_{4}, a_{5}\}$ is a uniform set and let $a_{1}<a_{2}<a_{3}<a_{4}<a_{5}$. Considering the set $A \setminus\{a_{1}\}$ we see that either $a_{2}+a_{5}=a_{3}+a_{4}$ or $a_{2}+a_{3}+a_{4}=a_{5}$. Considering the set $A \setminus\{a_{2}\}$ 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 uniform sets with 5 elements.