Gijswijt’s sequence throughout the integers

We define the Gijswijt’s sequence as follows :

$$\left\{ \begin{array}{ll} x_0=1,\\[2mm] x_{n+1} = 10x_n+ \displaystyle \max \left\{ k \in \mathbb{N} | \: \forall (i,j) \in [\![1,n]\!]^2, \: j>i \;\middle|\; \sum_{a=0}^{k}\left\lfloor\frac{x_n}{10^{n-j}}\right\rfloor\,10^{n-j+a} - \left\lfloor \frac{x_n}{10^{n-i}} \right\rfloor \,10^{n-i+a} = x_n -\left\lfloor \frac{x_n}{10^{n-k}} \right\rfloor 10^{n-k} \right\} \end{array} \right.$$

In other words, for a term $x_n$, look at its longest suffix that consists of repeated copies of the same block. The next term $x_{n+1}$ is the $x_n$ with number of copies of that block stick at its right end.

Show that every positive integer can be found in this sequence, i.e. that
$$\forall m \in \mathbb{N}, \exists n \in \mathbb{N} \text{, such that } max \left\{ k \in \mathbb{N} | \: \forall (i,j) \in[\![1,n]\!]^2, \: j>i \sum_{a=0}^{k}\left\lfloor\frac{x_n}{10^{n-j}}\right\rfloor\ 10^{n-j+a}-\left\lfloor \frac{x_n}{10^{n-i}} \right\rfloor \,10^{n-i+a} = x_n -\left\lfloor \frac{x_n}{10^{n-k}} \right\rfloor 10^{n-k} \right\} =m$$