【Question 13】If the product of two natural numbers, when divided by 9, leaves a remainder of 1, we say that these two natural numbers are “mod 9 reciprocals.” For example, $2 \times 5=10$, when divided by 9 leaves a remainder of 1, so 2 and 5 are “mod 9 reciprocals”; $1 \times 1=1$, so the “mod 9 reciprocal” of 1 is itself. Clearly, if a natural number has a “mod 9 reciprocal,” then its reciprocal is not unique; for example, 10 is another “mod 9 reciprocal” of 1. Determine whether 1, 2, $3,4,5,6,7,8$ have “mod 9 reciprocals,” and write down the numbers that have “mod 9 reciprocals” and their corresponding smallest “mod 9 reciprocals” respectively. $\qquad$
When $a$ and $b$ satisfy $a \times b = 9n + 1$, $a$ and $b$ are "mod 9 inverses" ($a$, $b$, $n$ are all natural numbers). When $a = 1$, the smallest "mod 9 inverse" is 1.
When $a = 2$, the smallest "mod 9 inverse" is 5.
When $a = 3$, $a \times b = 3b$, and $3b$ modulo 9 can only be 0, 3, or 6, so 3 has no "mod 9 inverse".
When $a = 4$, the smallest "mod 9 inverse" is 7.
When $a = 5$, the smallest "mod 9 inverse" is 2.
When $a = 6$, $a \times b = 6b$, and $6b$ modulo 9 can only be 0, 3, or 6, so 6 has no "mod 9 inverse".
When $a = 7$, the smallest "mod 9 inverse" is 4.
When $a = 8$, the smallest "mod 9 inverse" is 8.