Existence of 2006 distinct natural numbers

Question

Does there exists $2006$ distinct natural numbers such that the sum of any two divides the sum of all the given numbers?

Answer 1

The answer is no, and for any $n$, not just $n=2006$.

Suppose $a_1<a_2<\cdots<a_n$ be a such numbers, with the desired property. Let $S=\sum_{k=1}^n a_k$. Then observe that, $a_n+a_{n-1},a_n+a_{n-2},\cdots,a_n+a_1$ all divide $S$, and $$ \frac{S}{a_n+a_1}>\frac{S}{a_n+a_2}>\cdots>\frac{S}{a_n+a_{n-1}} \geq 2. $$ Hence, $$ \frac{S}{a_n+a_1}\geq n+1 \implies a_1+\cdots+a_{n-1}\geq na_n+(n+1)a_1, $$ which is a clear contradiction.