Pa-Lin-Dromes

Since x is 3 digit and x + 192 is four digit,

1000 - 192 = 808, the minimum possible value of x.

999 + 192 = 1193, the maximum possible value of x + 192.

To understand why I did this, note that one number is 3 digit, and the other is 4.

We're as such looking for x that when 192 is added, we get a 4 digit number.

It is obvious that all palis in the 800's will not work.

This is because 2 ends 192, and 8 ends any 8a8 palindrome. Thus the sum will end in 0, which cannot be.

Listing the 9b9 palis, we see 919 can work.

Because 919 + 192 = 1111.

Aliter,

since 1193 is the largest possible value, the closest pali to it is 1111.

1111 - 192 = 919

which is a pali, yay!

9 + 1 + 9 = 19.

So the question says that $x$ is a 3 digit palindrome , which means that it can be represented as a number $aba$ where $a$ and $b$ belong to the set $(1,2,3 .... 9 )$ and $b$ can also be $0$ .

So $x = a\times 100 + b\times 10 + a$

The question then says that $x +192$ is also a palindrome but it is a 4 digit number . Let's call it $y$ .

So,

$\begin{aligned} y&= x+192 \\ y&= (a\times 100 + b\times 10 + a) + (1\times100+9\times10+2) \\ y&=(a+1)\times100+(b+9)\times10+(a+2) .... (1)\\ \end{aligned}$

Now what we know is that any 4 digit number, is always of the form $p\times1000+q\times100+r\times10+s$ where $p,q,r,s$ belong to the set $(1,2,3....,9)$ and $q,r$ can be $0$ as well. So $y$ should also be of that form.

Which means that we have to introduce at least one number in thousands in the equation $(1)$ .

There's no hope for $(a+2)$ to become a number in thousands so we strike off that possibility. $(b+9)$ too can't help since the highest $b$ can be is $9$ and even then we don't get a number in thousands. $(a+1)$ is our only hope now, and it does the work! If $a=9$ we get $10\times100 = 1000$ .

Simplifying $y$ after substituting $a =9$ we get

$\begin{aligned} y&= 1000 + (b+9)\times10 + 11 \\ y&= 1101 + 10b \\ \end{aligned}$

At this point, its gets totally obvious that $b=1$ . No other value of $b$ makes $y$ a palindrome.

Hence, we conclude that $x = 919$ and $y = 1111$ .The sum of the digits of $x$ is $9+1+9 = 19$ .