14. solution: if n≤6, take m? 1, 2, …, 7, according to the pigeon hole principle, there must be12na, a, …, a.

A positive integer m in is the magic number of the common * * of I and j (1 ≤ I < j ≤ 7), that is, 7|( 10M? I),

7|( 10M？ J) .. There are 7|( j? I), but 0 < j? I≤6, contradictory.

Therefore, n ≥ 7.

When 1 2 n a, a, …, a is 1, 2, …, 7, let any positive integer m be k-bit.

Number (k is a positive integer). So 10ki? M (me? 1, 2, …, 7) Dividing the remainder by 7 is not the same. Otherwise,

There is a positive integer I, j (1 ≤ I < j ≤ 7), which satisfies 7|[( 10) (10 )] k k j? m？ Me? M, which is 7| 10 () k j? I

Thus 7| (j? I), contradiction.

So there must be a positive integer i (1≤i ≤7), which makes 7|( 10) ki? M, that is, the magic number with I as m.

Therefore, the minimum value of n is 7.