A palindrome is a number that is the same when read forwards and backwards, such as $43234$. Find the number of $4$-digit palindromes that are divisible by $13.$
Let the palindrome in question be $abba$, where $a$ and $b$ are digits. By Rule 1 of Divisibility by $13$, the palindrome is divisible by 13 if and only if $10a+b \equiv 10b+a \pmod{13},$ which gives us $9(a-b) \equiv 0 \pmod{13}.$
Since $9$ and $13$ are relatively prime, it follows that $a - b$ is divisible by 13.
Since $a$ and $b$ are digits, $a-b$ can only equal $-12,$ $-11$, $\dots$, $-1$, $0$, $1$, $\dots$, or $11.$
Thus, there are $11 - (-12) + 1 = \boxed{24}$ palindromes divisible by 13.
To solve this problem, we can use the divisibility rule for $13.$
The divisibility rule for $13$ states that a number is divisible by $13$ if and only if the sum of its digits, subtracting the sum of the first, third, fifth, and so on odd-placed digits from the sum of the second, fourth, sixth, and so on even-placed digits, is divisible by $13.$
Since the palindromes have an even number of digits, we can modify the rule slightly. The number is divisible by $13$ if and only if the sum of its digits, subtracting half of the sum of the first, third, fifth, and so on odd-placed digits from half of the sum of the second, fourth, sixth, and so on even-placed digits, is divisible by $13.$
Since the number is a $4$-digit palindrome, its digits can be represented as $\overline{ABBA},$ where $A$ and $B$ are single digits. We know that $A$ cannot be $0,$ since a $4$-digit number must start with a nonzero digit. Additionally, $A$ cannot be $6$ or greater, since $13 \times 6000 = 78000,$ which is a $5$-digit number and not a $4$-digit palindrome. Therefore, $A$ can be $1, 2, 3, 4,$ or $5.$
If $A = 1,$ then $B = 3, 6,$ or $9,$ since the resulting number must be divisible by $13.$
If $A = 2,$ then $B = 7,$ since the resulting number must be divisible by $13.$
If $A = 3,$ then $B = 4,$ since the resulting number must be divisible by $13.$
If $A = 4,$ then $B = 1, 5,$ or $9,$ since the resulting number must be divisible by $13.$
If $A = 5,$ then $B = 8,$ since the resulting number must be divisible by $13.$
Thus, there are $3$ choices for $B$ when $A = 1,$ $1$ choice for $B$ when $A = 2,$ $1$ choice for $B$ when $A = 3,$ $3$ choices for $B$ when $A = 4,$ and $1$ choice for $B$ when $A = 5.$
Therefore, there are $3 + 1 + 1 + 3 + 1 = \boxed{9}$ different $4$-digit palindromes that are divisible by $13.$
To find the number of $4$-digit palindromes that are divisible by $13,$ we can go through all the possible palindromes and check divisibility by $13.$
Let's consider an example of a $4$-digit palindrome $ABBA.$ Notice that any $4$-digit palindrome can be written as $ABBA = 1000A + 100B + 10B + A = 1001A + 110B.$
For it to be divisible by $13,$ we need $1001A + 110B$ to be divisible by $13.$
Now let's break it down:
\begin{align*}
1001A + 110B &\equiv 0 \pmod{13} \\
\Rightarrow 11A + 10B &\equiv 0 \pmod{13}\\
\Rightarrow 11A &\equiv -10B \pmod{13}\\
\Rightarrow A &\equiv -10B \cdot 11^{-1} \pmod{13}.
\end{align*}
Since we are looking for $4$-digit palindromes, $A$ and $B$ can vary from $0$ to $9.$
To find the modular inverse of $11$ modulo $13,$ we can use the Extended Euclidean Algorithm:
\begin{align*}
13 &= 1 \cdot 11 + 2 \\
11 &= 5 \cdot 2 + 1 \\
2 &= 2 \cdot 1 + 0.
\end{align*}
Now, working back up:
\begin{align*}
1 &= 11 - 5 \cdot 2 \\
&= 11 - 5(13 - 1 \cdot 11) \\
&= 6 \cdot 11 - 5 \cdot 13.
\end{align*}
So, $11^{-1} \equiv 6 \pmod{13}.$
Substituting into our earlier equation, we have:
\begin{align*}
A &\equiv -10B \cdot 11^{-1} \pmod{13} \\
&\equiv -10B \cdot 6 \pmod{13}.
\end{align*}
Now, we can try values of $B$ from $0$ to $9$ and see which ones yield an integer value of $A.$
For $B=0,$ we have $A \equiv 0 \pmod{13}.$
For $B=1,$ we have $A \equiv -10 \cdot 1 \cdot 6 \equiv -60 \equiv 5 \pmod{13}.$
For $B=2,$ we have $A \equiv -10 \cdot 2 \cdot 6 \equiv -120 \equiv 2 \pmod{13}.$
For $B=3,$ we have $A \equiv -10 \cdot 3 \cdot 6 \equiv -180 \equiv 11 \pmod{13}.$
For $B=4,$ we have $A \equiv -10 \cdot 4 \cdot 6 \equiv -240 \equiv 9 \pmod{13}.$
For $B=5,$ we have $A \equiv -10 \cdot 5 \cdot 6 \equiv -300 \equiv 7 \pmod{13}.$
For $B=6,$ we have $A \equiv -10 \cdot 6 \cdot 6 \equiv -360 \equiv 4 \pmod{13}.$
For $B=7,$ we have $A \equiv -10 \cdot 7 \cdot 6 \equiv -420 \equiv 1 \pmod{13}.$
For $B=8,$ we have $A \equiv -10 \cdot 8 \cdot 6 \equiv -480 \equiv 12 \pmod{13}.$
For $B=9,$ we have $A \equiv -10 \cdot 9 \cdot 6 \equiv -540 \equiv 10 \pmod{13}.$
So, there are $2$ pairs $(A, B)$ that satisfy the equation, namely $(0, 0)$ and $(1, 5).$
The total number of $4$-digit palindromes that are divisible by $13$ is $2,$ since we can arrange the pair $(0, 0)$ in $1$ way and the pair $(1, 5)$ in $2$ ways (as each of $A$ and $B$ can be the middle digits).
Therefore, the answer is $\boxed{2}.$