Let's assume that $\pi = \sum_{i=0}^{\infty}\frac{a_i}{10^i}.$ Also, let $n$ be an arbitrary natural number.
Is there $j,k,m \in \mathbb{N}, (j < k < m)$ such that $|b_1 - b_{2}| = n$ when $b_1 = \sum_{i=j}^{k}a_i10^{k-i}, b_2 = \sum_{i=k+1}^{m}a_i10^{m-i}$?
for example: 0.14(15)(92)... so for |92-15|=77 there are j,k,m as wanted.
the original question was:
can a number $x = \sum_{i=0}^{\infty}\frac{x_i}{10^i}$ such that $\forall n \in \Bbb N\ \exists j,k,m \in \mathbb{N}, (j < k < m)$ such that $|b_1 - b_{2}| = n$ when $b_1 = \sum_{i=j}^{k}a_i10^{k-i}, b_2 = \sum_{i=k+1}^{m}a_i10^{m-i}$ is a rational number?
i would like for now a solution only for the first part but any solution will be welcomed.
It is widely believed that the decimal expansion of $\pi$ contains every string of digits, see for example Does Pi contain all possible number combinations?
If one accepts this as true, your statement follows readily. However, to show this is considered as very difficult.
While your problem is somewhat simpler, I am afraid it is sufficiently similar that it is too difficult to be answered definitely at this point in time (and likely quite a bit of time in the future).