Lazim rolls two $24$-sided dice. From the two rolls, Lazim selects the die with the highest number. $N$ is an integer not greater than $24$. What is the largest possible value for $N$ such that there is a more than $50$% chance that the die Lazim selects is larger than or equal to $N$?
1. **Define the problem and the probability condition:**
   Lazim rolls two 24-sided dice and selects the die with the highest number. We need to find the largest integer \( N \) such that the probability of the highest number being at least \( N \) is more than 50%.

2. **Calculate the probability of the highest number being at least \( N \):**
   Let \( X \) and \( Y \) be the outcomes of the two dice rolls. We are interested in \( \max(X, Y) \geq N \).

3. **Calculate the probability of the highest number being less than \( N \):**
   The probability that both dice roll a number less than \( N \) is:
   \[
   P(\max(X, Y) < N) = P(X < N) \cdot P(Y < N)
   \]
   Since each die has 24 sides, the probability that one die rolls a number less than \( N \) is:
   \[
   P(X < N) = \frac{N-1}{24}
   \]
   Therefore,
   \[
   P(\max(X, Y) < N) = \left( \frac{N-1}{24} \right)^2
   \]

4. **Calculate the probability of the highest number being at least \( N \):**
   The probability that the highest number is at least \( N \) is:
   \[
   P(\max(X, Y) \geq N) = 1 - P(\max(X, Y) < N) = 1 - \left( \frac{N-1}{24} \right)^2
   \]

5. **Set up the inequality for the probability condition:**
   We need this probability to be more than 50%, so:
   \[
   1 - \left( \frac{N-1}{24} \right)^2 > 0.5
   \]
   Simplifying this inequality:
   \[
   \left( \frac{N-1}{24} \right)^2 < 0.5
   \]
   Taking the square root of both sides:
   \[
   \frac{N-1}{24} < \sqrt{0.5}
   \]
   \[
   N-1 < 24 \sqrt{0.5}
   \]
   \[
   N-1 < 24 \cdot \frac{\sqrt{2}}{2}
   \]
   \[
   N-1 < 24 \cdot \frac{1.414}{2}
   \]
   \[
   N-1 < 24 \cdot 0.707
   \]
   \[
   N-1 < 16.968
   \]
   \[
   N < 17.968
   \]
   Since \( N \) must be an integer, the largest possible value for \( N \) is 17.

The final answer is \( \boxed{17} \)