Obviously, $m$ and $n$ cannot be even numbers, because even number is not factor of an odd number. If $m\geq n$ and $k$ is odd: Suppose we have some $m,n$ that has minimal sum $m+n$ with $m\geq n$. By writing it out, where $k\in\mathbb Z^+$, $$(2m-1)m+(2n-1)n=kmn$$Then, $$2m^2-m(1+kn)+2n^2-n=0$$$$x^2-0.5x(1+kn)+n^2-0.5n=0$$One solution is $$x_1=m$$and the second is by Viete's formulae, $x_2=0.5(1+kn)-x_1=\frac{n^2-0.5n}{x_1}$. This is integer by the first one and positive by the second one. By our assumption, we need $x_2\geq x_1$, which means $n^2\leq x_1^2\leq n^2-0.5n$, which is impossible. No solutions, if $k$ is an odd number. If $k$ is even, then obviously, both fractions are odd numbers (and thus their product is odd, $l=2a+1$) and then $$\frac{4mn+1-2m-2n}{mn}=l.$$Thus, where $t$ is odd, since odd-even is odd, we get that, $mn\mid 2m+2n-1\implies 2m+2n-1=tmn\implies 2n-1=m(tn-2)\implies m=\frac{2n-1}{tn-2}$. $t$ can be only $1$ or $3$, since $t$ is odd and if $t\geq 5$ we get that $tn-2>2n-1\Longleftrightarrow (t-2)n\geq 3n >1$. If $t=1$, then $m=\frac{2n-1}{n-2}=\frac{2(n-2)+3}{n-2}=2+\frac{3}{n-2}\implies n=3 \text{ or } n=5$. Thus, $(m,n)=(3,5)$ and $(m,n)=(5,3)$. If $t=3$, then $m=\frac{2n-1}{3n-2}\implies \text{ if } n>1, \text{ then } 3n-2>2n-1\Longleftrightarrow n>1$. Thus if $n=1$, we get that $m=1$ and the solution is $(m,n)=(1,1)$. Answer. $\boxed{\text{Only pairs are } (m,n)=(1,1), (m,n)=(3,5) \text{ and } (m,n)=(5,3) .}$

$m|2n-1 \Rightarrow m|2m+2n-1$ $n|2m-1 \Rightarrow n|2n+2m-1$ Note that since $m|2n-1$ we have $\gcd (m,n)=1$ so $mn|2m+2n-1$.Assume that $m \ge n$ if $n \ge 4$ then $mn \ge 4n>2m+2n-1$ so $n \le 3$.else is just an obvious case check.

As n|2m-1 and m|2n-1 WLOG ,ASSUME THAT n>=m Now as n|2m-1 and n|2n Then n|2(m+n)-1 Similarly m|2n-1 and m|2m Then m|2(m+n)-1 Simply by divisbility rule that a|b and a|c then a|b+c Now as 2(m+n)-1 is same mod m and n and n>=m Then we can build up two cases Case 1 m|n n=mk But we can see clearly that k<2 Now k=1 does not satisfy because gcd(m,2m-1)=1 Hence this case fails only satisfies when m=1 Hence n=1 Case 2 Crucial case as they both are different we can say they divide same number then the product of mn =2(m+n)-1 You can easily by some example such as 7|21 3|21 but 3 does not divide 7 but 3×7=21 Hence this is now a simple diophantine factorizing it's as (m-2)(n-2)=3 Giving n=5 ,m=3 But as this is symmetrical n=3,m=5 is one more solution