$N$ is a $50$-digit number (in decimal representation). All digits except the $26$th digit (from the left) are $1$. If $N$ is divisible by $13$, find its $26$-th digit.
1. Let's denote the 50-digit number \( N \) as \( N = 111\ldots1d111\ldots1 \), where \( d \) is the 26th digit from the left, and all other digits are 1. We need to find \( d \) such that \( N \) is divisible by 13.

2. We can express \( N \) as:
   \[
   N = 10^{49} + 10^{48} + \ldots + 10^{25} + d \cdot 10^{24} + 10^{23} + \ldots + 10^0
   \]

3. To determine the divisibility by 13, we need to consider the properties of powers of 10 modulo 13. We know that:
   \[
   10^1 \equiv -3 \pmod{13}
   \]
   \[
   10^2 \equiv 9 \pmod{13}
   \]
   \[
   10^3 \equiv -1 \pmod{13}
   \]
   \[
   10^4 \equiv 3 \pmod{13}
   \]
   \[
   10^5 \equiv -4 \pmod{13}
   \]
   \[
   10^6 \equiv 1 \pmod{13}
   \]
   This cycle repeats every 6 powers.

4. We can use this cycle to simplify the expression for \( N \) modulo 13. Grouping the terms in sets of 6:
   \[
   N = (10^{49} + 10^{48} + 10^{47} + 10^{46} + 10^{45} + 10^{44}) + (10^{43} + 10^{42} + 10^{41} + 10^{40} + 10^{39} + 10^{38}) + \ldots + (10^{5} + 10^{4} + 10^{3} + 10^{2} + 10^{1} + 10^{0})
   \]
   Each group of 6 terms sums to:
   \[
   10^5 + 10^4 + 10^3 + 10^2 + 10^1 + 10^0 \equiv -4 + 3 - 1 + 9 - 3 + 1 \equiv 5 \pmod{13}
   \]

5. There are 8 such groups of 6 terms in the 50-digit number, plus the term \( d \cdot 10^{24} \). Therefore:
   \[
   N \equiv 8 \cdot 5 + d \cdot 10^{24} \pmod{13}
   \]
   \[
   N \equiv 40 + d \cdot 1 \pmod{13}
   \]
   \[
   N \equiv 40 + d \pmod{13}
   \]
   \[
   N \equiv 40 + d \equiv 0 \pmod{13}
   \]

6. Solving for \( d \):
   \[
   40 + d \equiv 0 \pmod{13}
   \]
   \[
   d \equiv -40 \pmod{13}
   \]
   \[
   d \equiv 39 \pmod{13}
   \]
   \[
   d \equiv 0 \pmod{13}
   \]
   \[
   d \equiv 9 \pmod{13}
   \]

The final answer is \( \boxed{9} \).