A divisibility problem

There are two groups of four-digit numbers divisible by $11$ that fulfill the conditions of values of $a$ and $b$ . The first group is consisted of palindromes. Palindromes are numbers that are read the same from left to right and from right to left. Here is a proof of this claim, taking for example a number with a form of $\overline{abba}$

$\overline {abba}$ $=$ $1000\times a + 100\times b + 10\times b + a$ $=$ $1001a + 110b$

Because $11|1001$ and $11|110$ $=>$ $11|\overline{abba}$

Palindromes come in the following forms:

• $\overline{aaaa}$
• $\overline{bbbb}$
• $\overline{abba}$
• $\overline{baab}$

The other group consists of numbers that are in forms of:

• $\overline{aabb}$
• $\overline{bbaa}$

And here is the proof for this claim, taking for example a number with a form of $\overline{aabb}$ :

$\overline {aabb}$ $=$ $1000\times a + 100\times a + 10\times b + b$ $=$ $1100a + 11b$

Because $11|1100$ and $11|11$ $=>$ $11|\overline{aabb}$

There is a total of $153$ numbers that fulfill all of the requirements above.