Let the ID-number be $d_1d_2\ldots d_{16}$, where each $d_i$ is a digit ($0$ to $9$). By condition (i), $d_1d_2d_3d_4 = 2024$ is fixed. So, the remaining $12$ digits $d_5, d_6, \ldots, d_{16}$ can be chosen.

Let $S$ be the set of all possible $12$-digit strings (from $0$ to $9$ in each position), i.e., $|S| = 10^{12}$.

We are to select a subset $A \subset S$ such that for any two distinct elements $x, y \in A$, $x$ and $y$ differ in at least $2$ positions (digits).

We claim that the maximal possible size of $A$ is $10^{11}$.

Proof:

Let us construct such a set $A$ of size $10^{11}$.

Let $A$ be the set of all $12$-digit numbers whose last digit is $0$. That is, $d_{16} = 0$. There are $10^{11}$ such numbers, since $d_5, d_6, \ldots, d_{15}$ can be arbitrary ($11$ digits), and $d_{16} = 0$.

Now, for any two distinct elements $x, y \in A$, they must differ in at least one of the first $11$ digits (since the last digit is always $0$). But since they are distinct, there is at least one position among $d_5$ to $d_{15}$ where they differ. However, we need to ensure that they differ in at least $2$ digits.

So, instead, let us consider the set $A$ of all $12$-digit numbers whose last two digits are $00$. That is, $d_{15} = d_{16} = 0$. There are $10^{10}$ such numbers.

But the problem asks for $10^{11}$.

Let us generalize. Consider the set $A$ of all $12$-digit numbers where $d_5$ is fixed (say $0$), and the remaining $11$ digits are arbitrary. Then $|A| = 10^{11}$.

For any two distinct elements $x, y \in A$, they differ in at least one of $d_6$ to $d_{16}$. Since $d_5$ is fixed, the only way $x$ and $y$ can differ is in $d_6$ to $d_{16}$. But since they are distinct, there is at least one position where they differ. However, it is possible that they differ in only one digit, which does not satisfy the condition.

Therefore, we need to find the largest possible subset of $S$ such that any two elements differ in at least $2$ digits.

This is a classical coding theory problem: the largest possible size of a code of length $12$ over an alphabet of size $10$ with minimal distance $2$.

The answer is given by the following construction:

Let $A$ be the set of all $12$-digit numbers with no two the same, and such that for each $x \in S$, at most one element of $A$ is at Hamming distance $0$ or $1$ from $x$.

Let us count the number of $12$-digit numbers at Hamming distance $0$ or $1$ from a fixed $x$:
- Hamming distance $0$: $x$ itself.
- Hamming distance $1$: For each of the $12$ positions, there are $9$ choices (since each digit can be changed to $9$ other digits), so $12 \times 9 = 108$.

So, for each $x$, there are $1 + 108 = 109$ numbers at Hamming distance $0$ or $1$ from $x$.

If $A$ is a code with minimal distance $2$, then the balls of radius $1$ around each codeword are disjoint and cover $|A| \times 109$ numbers. Since the total number of $12$-digit numbers is $10^{12}$, we have:
$$|A| \leq \frac{10^{12}}{109} < 10^{11}$$

But the problem claims that $10^{11}$ is possible.

Let us construct $A$ as follows:
Let $A$ be the set of all $12$-digit numbers where the sum of the digits is congruent to $0$ modulo $10$.

For any two distinct elements $x, y \in A$, their digit sums are both $0$ modulo $10$. If $x$ and $y$ differ in only one digit, say at position $i$, then the sum of digits of $x$ and $y$ would differ by a nonzero amount, so their sums modulo $10$ would be different, which is a contradiction. Therefore, any two elements of $A$ differ in at least $2$ digits.

How many such numbers are there? For any choice of the first $11$ digits, the last digit is determined modulo $10$ to make the sum $0$ modulo $10$. So, $10^{11}$ such numbers.

Therefore, the maximal possible number is $10^{11}$.

Thus, the answer is $10^{11}$.