Generalization of Palindromic number Consider a number $n > 0$ in base $b ≥ 2$, where it is written in standard notation with $k+1$ digits $a_i$ as:
$${\displaystyle n=\sum _{i=0}^{k}a_{i}b^{i}}$$
with, as usual, $0 ≤ a_i < b$ for all $i$ and $a_k ≠ 0$. Then n is palindromic if and only if $a_i = a_{k-i}$ for all $i$. 
And defined $P_l$ as
$k+1$ is number of digits in $n$
$$ (\sum_{i=0}^{k}|a_{i}-a_{k-i}|)/2=l$$
Then
$$ n\in P_l$$
It means $P_0 =$  palindromic number 
Example
● $121 \in P_0$
● $357 \in P_4$
● $9 \in P_0$
Let's defined function $\lambda_{b,n} (j)$ as
$\lambda_{b,n} (j)$ is legal if natural number $n \in P_r$ with base $b$ converted in $n \in P_j$ through the iterative process of repeatedly reversing its digits and adding the resulting numbers (This process is sometimes called the 196-algorithm). so $r$ is depend on $b,n$
Note If n is Lychrel number then $\lambda_{b,n} (0)$ is not legal 
Example
Let's given $j=16$, $b=10$, $n=9519$
To check $\lambda_{10,9519} (16)$ is legal or not 
$9519\in P_4$ so
$ 9519+9159=18678$ again $18678+87681=106359$ where $106359 \in P_{16}$ 
Hence $\lambda_{10,9519} (16)$ is legal


Question
1) Can we prove $\lambda_{10,n} (1)$ is legal for all $ n\in \mathbb{N} $ 
2) Can we prove there exist legal $\lambda_{b,n} (j)$ for all $ n,j\in \mathbb{N} $ for some $b$


 A: This is a partial answer too long for a comment:


*

*I will explain why your first question is probably false and why we can't prove/disprove it.

*I will prove your other question is trivial for all $n$ such that:  ($j=0,1$) or if ($j$ divides $n\gt j^2$). Maybe someone else can work out the full proof If I don't get back to this.


Regarding your $1.$ question: "I don't think it is true and I do not think we can."
Your first question will be as hard as the Lychrel Numbers which is an open problem. We do not know for example if $196$ is truly a Lychrel Number. The sequence is just conjectured based on extensive computations, as we do not know how to solve these kind of problems (for base ten at least).
Because we do not know how to prove given $n$, that $\lambda_{10,n}(0)$ is "not legal".
And you are asking about $\lambda_{10,n}(1)$ which I don't see why would be any easier. I have ran a computation, and $n=101$ seems to be the smallest "not legal" number which would be a counter example to what you want proven. - But as I said, this is probably as hard to prove as $196$ being smallest "not legal" for  $\lambda_{10,n}(0)$.
The $n=101$ does not turn into an element of $P_1$ for $\gt 3000$ reversal steps, I haven't computed more.
The number of steps numbers $1,\dots, 100$ need to become $\in P_1$, in comparison:
7, 6, 2, 5, 1, 1, 3, 4, 12, 0, 5, 0, 4, 2, 2, 3, 3, 11, 1, 5, 0, 4, 0, 2, 3, 3, 11, 1, 4, 3, 4, 0, 2, 0, 3, 11, 1, 4, 1, 4, 2, 2, 0, 3, 0, 1, 4, 1, 3, 2, 2, 3, 3, 0, 1, 0, 1, 3, 2, 2, 3, 3, 11, 1, 0, 1, 0, 2, 2, 3, 3, 11, 1, 4, 1, 0, 2, 0, 2, 3, 11, 1, 4, 1, 3, 2, 0, 2, 0, 11, 1, 4, 1, 3, 2, 2, 2, 0, 10, 0


Proving your $2.$ question for trivial cases:
You are asking if every number $n\in\mathbb N$ will turn into element of $P_j$ for every $j$, for some number base $b$?
I will denote digits of $n$ in base $b$ as $(n_1,n_2,\dots)_b$.
If $r=j$, then we are done, since $n\in P_j$ before we even start reversing digits. 
I will find simplest of $b$ such that $r=j$  to prove the statement for given $j,n$ for trivial cases:


*

*This is true for $j=0$ for all $n$. Take $b\gt n$ so $n=(n)_b$.

*This is true for $j=1$ for all $n\gt 1$. Take $b=n$ so $n=(1,0)_b$. 

*For $j=1$, it works for $b=10$ for $n=1$. Same for all $n\lt 101$ as computed in $1.)$ question.

*This is true if $n\gt j^2$ and is divisible by $j$, we have $n=(j,0)_b$ for $b=n/j$. 
Given this, what is left to prove:

We are left to consider $n\le j^2$ and divisible by $j$, and $n\in\mathbb N$ not divisible by $j$, for $j\ge 2$ in both cases.
I will update the answer if I continue the proof. - I also encourage anyone reading this to try to continue the proof for non-trivial cases and post their own answer if they succeed.

