Konigsberg wrote: Find all positive integers $n$ for which there are distinct integers $a_1$, ..., $a_n$ such that $\frac{1}{a_1}+\frac{2}{a_2}+...+\frac{n}{a_n}=\frac{a_1+a_2+...+a_n}{2}$. It's rather easy to show that no solution exists for $n=1$ or $n=2$ A solution exists for $n\in\{3,4,5,6\}$ : $\frac 11+\frac 22+\frac 33=\frac{1+2+3}2$ $\frac 11+\frac 22-\frac 33+\frac 44=\frac{1+2-3+4}2$ $\frac 11+\frac 22-\frac 33-\frac 48+\frac 5{10}=\frac{1+2-3-8+10}2$ $\frac 11+\frac 22+\frac 33-\frac 44-\frac 5{10}+\frac 6{12}=\frac{1+2+3-4-10+12}2$ If a solution exists for a given $n$, a solution exists for $n+4$ : choose $k\in\mathbb N$ such that $k(n+1)>\max_{i\in[1,n]}|a_i|$ Then $(a_1,a_2,...,a_n,k(n+1),-k(n+2),-k(n+3),k(n+4))$ is a solution too. Hence the result : $\boxed{n\in[3,+\infty)}$