Problem 1839

$$\sum_{i=1}^{n} \frac{1}{b_i} = k_n \sum_{i=1}^{n} \frac{1}{a_i}$$
by subtracting these equalities for $n, n + 1$ we have:
$$\frac{1}{b_{n+1}} = k_{n+1} \frac{1}{a_{n+1}} + (k_{n+1} - k_n) \sum_{i=1}^{n} \frac{1}{a_i}$$
Choose a big enough $m$, such that for $n > m$ we have $b_n > a_0$, (this is possible because $b_i$ are distinct natural numbers). Then we have $\frac{1}{b_{n+1}} < \sum_{i=1}^{n} \frac{1}{a_i}$.
Therefore $k_{n+1} \le k_n$.
Whence, for big enough $i$, the sequence $k_i$ is decreasing, hence they are constant after some point. If $k_n = k_{n+1}$ then because $\gcd(a_{n+1}, a_n) = 1$ we have $k_n = k_{n+1} = 1$. So we have $a_n = b_n$ for large enough $n$.
Assume that for $n > N$, $a_n = b_n$, $k_n = 1$, by subtracting the equations
$$\sum_{i=1}^{N+1} \frac{1}{b_i} = \sum_{i=1}^{N+1} \frac{1}{a_i}, \quad a_{N+1} = b_{N+1}$$
we get $\sum_{i=1}^{N} \frac{1}{b_i} = \sum_{i=1}^{N} \frac{1}{a_i}$ and so $k_N = 1$. Now we have
$$\frac{1}{b_N} = \frac{1}{a_N} + (1 - k_{N-1}) \sum_{i=1}^{N-1} \frac{1}{a_i}$$
the $a_i$ are increasing so if $k_{N-1} \neq 1$ the right hand side becomes negative which is a contradiction. Hence we have $k_{N-1} = 1$ and by induction, we prove that $a_i = b_i$. ■