Autumn tournament

Set $b_i=2{\ell }_i$, all of them are even. At the first step we get the pairs $$A:=\{(p,q):(p,q)=(1,2{\ell }_1),{\ell }_1\in \mathbb{Z}.(1)\}.$$ At each subsequent step we expand the set $A$ by adding additional pairs $(p^{\prime },q^{\prime })$ defined as $$\left\{(p^{\prime },q^{\prime }):\frac{p^{\prime }}{q^{\prime }}=\frac{1}{2\ell +\frac{p}{q}},(p,q)\in A,\ell \in \mathbb{Z}\right\}(2).$$ By (2) we obtain $$p^{\prime }=q;q^{\prime }=2\ell q+p(3)$$ So, if $(p,q)$ is obtained in the process of expanding, we also add $(p^{\prime },q^{\prime })$ defined as in (3). We want to characterize all pairs $(p,q)$ that can be obtained in this way. Note that if $(p,q)=1$, then from (3) follows $(p^{\prime },q^{\prime })=1$. Let us prove that the set of pairs $(p,q)$ in question is $$A:=\{(p,q):p,q\in \mathbb{Z}\setminus \{0\},|p|<|q|,(p,q)=1,\text{ one of }p,q\text{ is even, the other is odd.}\}$$ Note that we ruled out $|p|=|q|$ because $\frac{p}{q}=\pm 1$ cannot be represented as wanted. The transformation described in (3) shows that we cannot step outside $A$. To prove that all pairs in $A$ can be generated, we follow a standard procedure - take $(p^{\prime },q^{\prime })\in A$ and search for $(p,q)\in A$ that generates $(p^{\prime },q^{\prime })$ via (3), but we want $(p,q)$ be "less" than $(p^{\prime },q^{\prime })$. That's how induction works. Solving (3) with respect to $(p,q)$ yields $$p=q^{\prime }-2\ell p^{\prime };q=p^{\prime }(4)$$ Clearly, we can choose $\ell \in \mathbb{Z}$ such that $|q^{\prime }-2\ell p^{\prime }|<|p^{\prime }|$ (since $p^{\prime }\ne 0$). Indeed, consider the points $q^{\prime }-2\ell p^{\prime }$ where $\ell$ runs through all integers. These points are at a distance $2p^{\prime }$ apart, and none of them hits $p^{\prime }$ because $(p^{\prime },q^{\prime })=1$. Hence, the point closest to 0 has magnitude less than $p^{\prime }$. So, we started from $(p^{\prime },q^{\prime })\in A$, and found $(p,q)\in A$ that generates $(p^{\prime },q^{\prime })$ and moreover, $q=p^{\prime },|p|<|p^{\prime }|<|q^{\prime }|$. Apparently $(p,q)$ is "less" than $(p^{\prime },q^{\prime })$, that is, $|p|<|p^{\prime }|,|q|<|q^{\prime }|$. Now, induction does the job. Obviously, $(1,2),(-1,2),(1,-2),(-1,-2)$ can be represented as continued fractions satisfying the statement. Assume that all $(p,q)\in A$ with $|p|\le N,|q|\le N$ can be represented like that. Take $(p^{\prime },q^{\prime })\in A,|q^{\prime }|=N+1$. As just shown, there exists $(p,q)\in A,|p|<|q|\le N$ such that $$\frac{p^{\prime }}{q^{\prime }}=\frac{1}{2\ell +\frac{p}{q}}.$$ The induction step is complete and the result follows. Note that it also follows that the representation of numbers in $A$ as continued fractions like that is unique. Indeed, take a pair $(p^{\prime },q^{\prime })\in A$. The pair $(p,q)\in A$ that satisfies (4) and $|p|<|p^{\prime }|$ is determined uniquely. Uniqueness follows easily.