Mongolian Mathematical Olympiad

Note that $(n+m-1)|(n+m-1{)}^2+2(n+m-1)=(n+m-1{)}^2+2mn$. By combining it with given condition we get $(n+m-1)|2mn$. Since none of $(n+m-1)$ and $(n+m+1)$ equals to $2$, greatest common divisor (GCD) of these numbers is not greater than $2$. If $GCD=2$ then $\frac{1}{2}(n+m-1)(n+m+1)|2mn$. Also if $GCD=1$ then $(n+m-1)(n+m+1)|2mn$. From here we deduce that $2mn\ne 0$, $\frac{1}{2}(n+m-1)(n+m+1)\le 2mn$ and $(n+m{)}^2-1\le 4mn\Rightarrow (n+m{)}^2\le 1$ $(\ast )$ Since $n,m$ symmetrical, we may assume $n\ge m$. Considering the $(\ast )$, we get either $n=m$ or $n=m+1$

a. Let $n=m$. By the given condition $2m+1|2m^2$ and $(2m+1,m)=1$, $(2m+1,2)=1$. It is obvious that there are no such numbers.

b. Let $n=m+1$. By the given conditions $$\{\begin{array}{l}(n+m+1)|2mn\\ (n+m-1)|n^2+m^2-1\end{array}\iff \{\begin{array}{l}2(m+1)|2mn(m+1)\\ 2m|2mn(m+1)\end{array}$$ and we conclude that any natural $m$ satisfies the conditions.

Thus $(n,m)$ is pair of successive natural numbers.