A four-digit palindrome number is of the form $1000a+100b+10b+a=1001a+110b=11(91a+10b)$ so is divisible by $11$. So a balanced number that is the sum of two palindrome numbers must be divisible by $11$, meaning $a+c=b+d$ or $a+c=b+d\pm11$. If $a+c=b+d\pm{11}$ then with $a+b=c+d$ we have $2a=2d\pm11$. This is impossible as $11$ is not even. So $a+c=b+d$; with $a+b=c+d$ this gives $a=d$, $b=c$. So $\overline{abcd}$ itself is palindromic. To be the sum of two palindrome numbers however $a$ can’t be $1$ so this gives $2002$, $2112$, $\ldots$, $9999$.

MillMaths, your solution is incomplete, because 2011 = 2002+9 and 2002, 9 are both palindrome numbers.

The problem here is that MillMaths assumed the number is the sum of two four-digit palindromes. Solving in general can be done I bet in a similar way to MillMaths' solution, but it's just a godawful bash and I have no time to do this.

If the palindromes have 3, 2 or 1 digit, the problem is more dificult