Example 26 (Problem from the 30th Russian Mathematical Olympiad) Let $\left\{N_{1}, \cdots, N_{k}\right\}$ be an array of five-digit numbers (in decimal) such that any five-digit number with digits in non-decreasing order has at least one digit in the same position that is the same as a digit in one of the numbers $N_{1}, \cdots, N_{k}$. Find the minimum possible value of $k$.
The smallest value sought is 2. The reason is as follows:
Au
Prohibited
Lian
Mathematics
An array with the described property cannot consist of a single five-digit number. In fact, for any five-digit number \( N = \overline{a b c d e} \), there exists a digit \( g \) different from \( a, b, c, d, e \), thus, no digit of \( N \) is the same as \( G = \overline{g g g g g} \).
The following proof shows that the array consisting of the numbers \( N_{1} = \overline{13579} \) and \( N_{2} = \overline{12468} \) satisfies the condition.
Take any five-digit number \( A = \overline{a_{1} a_{2} a_{3} a_{4} a_{5}} \), where \( a_{1} \leqslant a_{2} \leqslant a_{3} \leqslant a_{4} \leqslant a_{5} \). If any digit of \( A \) does not match the digit in the same position of \( N_{1} \) and \( N_{2} \), then it must be that \( a_{5} \leqslant 7 \), and therefore, \( a_{4} \leqslant 7 \). However, since \( a_{4} \) is different from the tens digit of \( N_{1} \) and the tens digit of \( N_{2} \), it follows that \( a_{4} \leqslant 5 \), hence \( a_{3} \leqslant 5 \). Since \( a_{3} \) is different from the hundreds digit of \( N_{1} \) and the hundreds digit of \( N_{2} \), it follows that \( a_{3} \leqslant 3 \), hence \( a_{2} \leqslant 3 \). Continuing this reasoning, it follows that \( a_{2} \neq 3 \) and \( a_{2} \neq 2 \), thus, it must be that \( a_{1} = 1 \), and therefore, its leading digit is the same as the leading digit of \( N_{1} \) (and also of \( N_{0} \)).