Find the largest $2$-digit number $N$ which is divisible by $4$, such that all integral powers of $N$ end with $N$.
1. We need to find the largest 2-digit number \( N \) which is divisible by 4, such that all integral powers of \( N \) end with \( N \). This means \( N^k \equiv N \pmod{100} \) for all \( k \geq 1 \).

2. To determine the last two digits of a number, we take the number modulo 100. Therefore, we need:
   \[
   N^2 \equiv N \pmod{100}
   \]
   This simplifies to:
   \[
   N^2 - N \equiv 0 \pmod{100}
   \]
   Factoring out \( N \), we get:
   \[
   N(N - 1) \equiv 0 \pmod{100}
   \]

3. Since \( N \) is a 2-digit number divisible by 4, we have:
   \[
   N \equiv 0 \pmod{4}
   \]

4. We also need \( N(N - 1) \equiv 0 \pmod{100} \). This means \( N \) or \( N - 1 \) must be divisible by 25 because 100 = 4 * 25 and we already have \( N \equiv 0 \pmod{4} \).

5. The largest two-digit multiple of 25 is 75. However, since \( N \) must be divisible by 4, we consider \( N - 1 \) to be divisible by 25. Therefore:
   \[
   N - 1 = 75 \implies N = 76
   \]

6. We verify that \( N = 76 \) satisfies all conditions:
   - \( 76 \equiv 0 \pmod{4} \)
   - \( 76 - 1 = 75 \equiv 0 \pmod{25} \)
   - \( 76^2 = 5776 \equiv 76 \pmod{100} \)

Thus, \( N = 76 \) is the largest 2-digit number divisible by 4 such that all integral powers of \( N \) end with \( N \).

The final answer is \( \boxed{76} \).