Solution:
Let $m_{1}, n_{1}, m_{2}, n_{2}$ be integers satisfying $m_{1} n_{2}-m_{2} n_{1}= \pm 1$. By changing the signs of $m_{2}, n_{2}$ if need be, we may assume that
$$m_{1} n_{2}-m_{2} n_{1}=1$$
If $m_{3}, n_{3}$ are integers satisfying $m_{1} n_{3}-m_{3} n_{1}= \pm 1$, again we may assume (by changing their signs if necessary) that
$$m_{1} n_{3}-m_{3} n_{1}=1$$
So $m_{1}\left(n_{2}-n_{3}\right)=n_{1}\left(m_{2}-m_{3}\right)$.
As $m_{1}, n_{1}$ are relatively prime, $m_{1}$ divides $m_{2}-m_{3}$; say, $m_{2}-m_{3}=m_{1} a$ for some integer $a$. Thus, we get $n_{2}-n_{3}=n_{1} a$. In other words,
$$m_{3}=m_{2}-m_{1} a, \quad n_{3}=n_{2}-n_{1} a$$
Now, if $m_{2} n_{3}-n_{2} m_{3}= \pm 1$, we get
$$\pm 1=m_{2}\left(n_{2}-n_{1} a\right)-n_{2}\left(m_{2}-m_{1} a\right)=\left(m_{1} n_{2}-m_{2} n_{1}\right) a=a$$
Thus, $m_{3}=m_{2}-m_{1} a=m_{2} \pm m_{1}, \quad n_{3}=n_{2}-n_{1} a=n_{2} \pm n_{1}$.
Now if we were to have another pair of integers $m_{4}, n_{4}$ such that
$$m_{1} n_{4}-n_{1} m_{4}= \pm 1$$
we may assume that $m_{1} n_{4}-n_{1} m_{4}=1$. As seen above, $m_{4}=m_{2} \mp m_{1}, \quad n_{4}=n_{2} \mp n_{1}$. But then
$$m_{3} n_{4}-n_{3} m_{4}=\left(m_{2} \pm m_{1}\right)\left(n_{2} \mp n_{1}\right)-\left(n_{2} \pm n_{1}\right)\left(m_{2} \mp m_{1}\right)= \pm 2$$
Therefore, there can be only 3 pairs of such integers.

Alternate Solution:
It is clear that $r$ can be 3 due to the valid solution $m_{1}=1, n_{1}=1, m_{2}=1, n_{2}=2, m_{3}=2, n_{3}=3$.
If possible, let $r>3$. We observe that:
$$\begin{aligned} & m_{1} n_{2} n_{3}-m_{2} n_{1} n_{3}= \pm n_{3} \\ & m_{2} n_{3} n_{1}-m_{3} n_{2} n_{1}= \pm n_{1} \\ & m_{3} n_{1} n_{2}-m_{1} n_{3} n_{2}= \pm n_{2} \end{aligned}$$
Adding, we get $\pm n_{1} \pm n_{2} \pm n_{3}=0$; which forces at least one of $n_{1}, n_{2}, n_{3}$ to be even; WLOG let $n_{1}$ be even.
Repeating the argument for indices $2,3,4$, we deduce that at least one of $n_{2}, n_{3}, n_{4}$ is even; WLOG let $n_{2}$ be even. This leads to a contradiction, since $\left|m_{1} n_{2}-m_{2} n_{1}\right|=1$ cannot be even. Hence $r>3$ is not possible, as claimed.