no need apply brains

You have a number  $\overline { abc }$ and the condition $a+b+c=4b$

solve for b you get:

$a+b+c=4b\\ a+c=4b-b\\ a+c=3b\\ (1):\frac { a+c }{ 3 } =b$

since $\overline { abc }$ is divisible by 3, then by divisibility rules:

$a+b+c=3Z$ where $Z$ is an integer

substitute the value of $b$ from $(1)$ and simplify:

$a+\frac { a+c }{ 3 } +c=3Z\\ 3a+a+c+3c=9Z\\ 4a+4c=9Z\\ a+c=\frac { 9 }{ 4 } Z$

since $a$ and $c$ are both integers, the RHS must be an integer, thus $Z$ can only be $4\quad or\quad 8$

if $Z=4$ then $a+c=9$ , there are $9$ possible answers for this: $\left\{ \left( 1,8 \right) ,(2,7),...,(9,0) \right\}$

if $Z=8$ then $a+c=18$ , and there is only $1$ answer $\left\{ (9,9) \right\}$

Add the number of possibilities for each case:

$9\quad +\quad 1\quad =\quad \boxed { 10 }$

Therefore the answer is $\boxed { 10 }$ and the numbers satisfying the conditions are: $\left\{ 138,237,336,435,534,633,732,831,930,969 \right\}$