We have $2^2$ pairs of numbers such that: $$\frac{1}{1\cdot 2}+\frac{1}{2\cdot 2}+\frac{1}{1\cdot 6}+\frac{1}{1\cdot 12}=1$$and $$(1+2+1+1)+(2+2+6+12)=3^3$$Suppose that for an integer $k\ge 2$ there exists $2^k$ pairs of natural numbers sucht that: $$\sum^{2^k}_{i=1} \frac{1}{a_ib_i}=1$$and $$\left(\sum^{2^k}_{i=1} a_i\right)+\left(\sum^{2^k}_{i=1} b_i\right)=3^{k+1}$$We define $(a^\prime_i,b^\prime_i)=(2a_i,b_i)$ and $(a^\prime_{2^k+i},b^\prime_{2^k+i})=(a_i,2b_i)$ for all $i\in\mathbb{N}\land i\le2^k$. So, we have $2^{k+1}$ pairs of numbers $(a^\prime_i,b^\prime_i)$ such that: $$\begin{aligned} \sum^{2^{k+1}}_{i=1} \frac{1}{a^\prime_ib^\prime_i} &=\left(\sum^{2^k}_{i=1} \frac{1}{a^\prime_ib^\prime_i}\right)+\left(\sum^{2^{k+1}}_{i=2^k+1} \frac{1}{a^\prime_ib^\prime_i}\right)\\ &=\left(\sum^{2^k}_{i=1} \frac{1}{2a_i\cdot b_i}\right)+\left(\sum^{2^k}_{i=1} \frac{1}{a^\prime_{2^k+i}b^\prime_{2^k+i}}\right)\\ &=\frac{1}{2}\left(\sum^{2^k}_{i=1} \frac{1}{a_ib_i}\right)+\left(\sum^{2^k}_{i=1} \frac{1}{a_i\cdot 2b_i}\right)\\ &=\frac{1}{2}\left(1\right)+\frac{1}{2}\left(\sum^{2^k}_{i=1} \frac{1}{a_ib_i}\right)\\ &=1 \end{aligned}$$and $$\begin{aligned} \left(\sum^{2^{k+1}}_{i=1} a^\prime_i\right)+\left(\sum^{2^{k+1}}_{i=1} b^\prime_i\right) &=\left(\sum^{2^{k}}_{i=1} a^\prime_i\right)+\left(\sum^{2^{k+1}}_{i=2^k+1} a^\prime_i\right)\\ &\phantom{=}+\left(\sum^{2^{k}}_{i=1} b^\prime_i\right)+\left(\sum^{2^{k+1}}_{i=2^k+1} b^\prime_i\right)\\ &=\left(\sum^{2^{k}}_{i=1} 2a_i\right)+\left(\sum^{2^k}_{i=1} a^\prime_{2^k+i}\right)\\ &\phantom{=}+\left(\sum^{2^k}_{i=1} b_i\right)+\left(\sum^{2^k}_{i=1} b^\prime_{2^k+i}\right)\\ &=\left(\sum^{2^k}_{i=1} 2a_i\right)+\left(\sum^{2^k}_{i=1} a_i\right)\\ &\phantom{=}+\left(\sum^{2^k}_{i=1} b_i\right)+\left(\sum^{2^k}_{i=1} 2b_i\right)\\ &=3\left[\left(\sum^{2^k}_{i=1} a_i\right)+\left(\sum^{2^k}_{i=1} b_i\right)\right]\\ &=3\cdot 3^{k+1}\\ &=3^{k+2} \end{aligned}$$And we are done. PD. If all the pairs must be different, the solution is wrong. Help me please

I think a little adjustment will work.can someone help?

We can choose (a_i+b_i,b_i) and (a_i,a_i+b_i) instead. It's easy to prove that they are different by induction.