Claim: The only $3$ consecutive odd numbers that are semi-primes are $5,7,9$. Proof: An integer can only be odd if it is a sum of an odd and an even integer. An odd integer can be a semi-prime only if it is a sum of $2$ and some odd prime. So, we can only have consecutive odd semi-primes if we also have consecutive odd numbers all of which are prime. It is easy to see that the longest chain of consecutive odd numbers that we could have such that they are all primes is $3,5,7$. (Anything longer than that, then at least one of the numbers would be a multiple of $3$ but greater than $3$, so it's not prime.) Therefore, the only $3$ consecutive odd numbers that are semi-primes are $5,7,9$.$\square$ I claim that the answer is $5$, which is achievable by $12,13,14,15,16$. To prove that the answer can only be at most $5$, suppose there exists $6$ consecutive natural numbers that are semi-primes. If we have $6$ consecutive natural numbers, then this means there must be $3$ consecutive odd numbers in that sequence. According to the claim, these three odd numbers could only be $5,7,9$. We have two cases: Case I. The $6$ numbers are: $4,5,6,7,8,9$ Case II. The $6$ numbers are: $5,6,7,8,9,10$. We can easily see that both of these don't work, since $6$ is not a semi-prime. This gives us a contradiction. Therefore, the greatest number of consecutive natural numbers that are semi-primes is $\boxed5$.