Answer: $N = 6$.

Let some digit, say $1$, appear exactly in $k$ numbers from $N$ given numbers. Hence, $1$ forms at most $3$ distinct pairs with the remaining $3$ digits of any of these $k$ numbers. Since the total number of all distinct pairs formed by $1$ and the other $7$ numbers ($2,3,\ldots,8$) is equal to $7$, we see that $3k \ge 7$. So $k \ge 3$. Therefore, each of the digits $1, 2, \ldots, 8$ must appear in at least $3$ numbers. Thus, the total number of all digits in $N$ numbers is greater than or equal to $8 \cdot 3 = 24$. But $N$ numbers contains exactly $4N$ digits. Therefore, $4N \ge 24$, so $N \ge 6$.

The following example shows that there are $6$ four-digit numbers satisfying the ...