Mongolian Mathematical Olympiad

Answer: $a_n=1+c(n-1)$ for fixed $c\ge 0$. It is clear that the above is a solution, so we prove there are no other solutions. The expression $$n-\frac{n{a}_n-m{a}_m+2a_m-1}{a_n+a_m-1}=\frac{n(a_m-1)+(m-2)a_m+1}{a_n+a_m-1}$$ is an integer for all $n,m\ge 1$. Let $c=a_2-1\ge 0$. Taking $m=2$, we see that $\frac{cn+1}{a_n+c}$ is a positive integer. If $c=0$, we have $a_n=1$ for all $n\ge 1$. So assume $c\ge 1$ and suppose that $cn+1$ is a prime. Clearly $a_n+c\ge 2$, therefore $a_n=1+c(n-1)$. Finally, $$(a_m-1)-c\frac{n(a_m-1)+(m-2)a_m+1}{cn+a_m-c}=\frac{a_m(a_m-1-c(m-1))}{cn+a_m-c}$$ is an integer for any $m\ge 1$. By Dirichlet's theorem, we may assume that $n$ is sufficiently large, so we must have $a_m=1+c(m-1)$.