9. (10 points) Among all the factors of a natural number, the factors that are divisible by 3 are 5 more than the odd factors, then the smallest natural number is $\qquad$ .
【Analysis】Consider performing prime factorization on this natural number, analyzing its prime factors and their counts, and then calculating this natural number.

【Solution】Since there exists a factor divisible by 3, this natural number must have 3 as a prime factor. Then, try to enumerate the factors divisible by 3, considering the prime number 2, from smallest to largest they are $2 \times 3=6, 3 \times 3=9, 2 \times 2 \times 3=12, 2 \times 3 \times 3=18, 2 \times 2 \times 2 \times 3=24, 2 \times 2 \times 3 \times 3=36$, up to this point, the condition is satisfied, thus the natural number obtained is $2 \times 2 \times 2 \times 3 \times 3=72$.

If considering that 72 is also its own factor, then 1 is also a factor of 72, still satisfying the condition that the number of factors divisible by 3 is 5 more than the number of odd factors. The problem is solved.

Therefore, the answer is 72.