As John's answer notes, every palindromic number with an even number of digits is divisible by $11$, because the alternating sum of its digits is zero (and thus a multiple of $11$).

For example, for $81233218$, we have: $$8-1+2-3+3-2+1-8 = 0,$$ and so $81233218$ is divisible by $11$.

The *reason* why this divisibility rule works is most easily seen using a bit of modular arithmetic.

Specifically, by definition, two numbers $a$ and $b$ are equivalent modulo $m$ (which we write as $a \equiv b \pmod m$) if and only if their difference $a-b$ is divisible by $m$. Thus, in particular, a number $n$ is divisible by $m$ if and only if $n \equiv 0 \pmod m$.

Now, the reason this definition is handy is because we can "do arithmetic under the modulus $m$": in particular, if $a \equiv a' \pmod m$ and $b \equiv b' \pmod m$, then $a+b \equiv a'+b' \pmod m$ and $ab \equiv a'b' \pmod m$. Thus, if we're only interested in the result of a calculation modulo some number $m$ (like, say, if we just want to know whether $n \equiv 0 \pmod{11}$ for some number $n$), and the calculation only involves operations that work equivalently "under the modulus" (such as addition and multiplication, and any combination thereof), then we can freely add or subtract multiples of $m$ from any intermediate values to make them more convenient (which often means "closer to zero").

In particular, from the definition it clearly follows that: $$10 \equiv -1 \pmod{11},$$ since $10 - (-1) = 10 + 1 = 11$. We also know that the numerical value of a base-$10$ number is mathematically obtained by multiplying its lowest digit with $1$, the second digit with $10$, the third digit with $10^2 = 100$, etc., and adding the results together. For example: $$81233218 = 10^7 \cdot 8 + 10^6 \cdot 1 + 10^5 \cdot 2 + 10^4 \cdot 3 + 10^3 \cdot 3 + 10^2 \cdot 2 + 10 \cdot 1 + 8.$$

But since $10 \equiv -1 \pmod{11}$, if we only want to calculate the value of the number modulo $11$, we can replace $10$ with $-1$ in this calculation! Thus: $$81233218 \equiv (-1)^7 \cdot 8 + (-1)^6 \cdot 1 + (-1)^5 \cdot 2 + (-1)^4 \cdot 3 + (-1)^3 \cdot 3 + (-1)^2 \cdot 2 + (-1) \cdot 1 + 8 \pmod{11}.$$

And, of course, we can easily see that the powers of $-1$ simply alternate between $-1$ and $1$, so this expression simplifies down to: $$81233218 \equiv -8 + 1 -2 + 3 - 3 + 2 - 1 + 8 = 0 \pmod{11}.$$

(Ps. The same trick also works in bases other than $10$, showing that any number palindromic in base $b$, with an even number of base-$b$ digits, is necessarily divisible by $b+1 = 11_b$.)