Introduction:
This problem requires us to find the number of natural numbers less than 200 such that (n-1)! is not divisible by n.

Solution:
To solve this problem, we need to consider the following cases:

Case 1: n is a prime number
- If n is a prime number, then using Wilson's theorem, we know that (n-1)! ≡ -1 (mod n)
- Therefore, n does not divide (n-1)!

Case 2: n is a composite number
- If n is a composite number, then it can be written as n = ab, where a and b are both integers greater than 1.
- Since a and b are both less than n, we know that (a-1)! and (b-1)! are both factors of (n-1)!.
- Therefore, n divides (n-1)! if and only if n divides either (a-1)! or (b-1)!.

Conclusion:
Using the above cases, we can conclude that:
- The number of natural numbers less than 200 such that (n-1)! is not divisible by n is equal to the number of prime numbers less than 200.
- To find this number, we can use the Sieve of Eratosthenes to find all prime numbers less than 200.
- Finally, we can count the number of prime numbers found to get our answer.

Answer:
There are 46 natural numbers less than 200 such that (n-1)! is not divisible by n.

All 46 Prime Number + 1 Exception, In which (n-1)!/n is not dividable

n=2, (2-1)! / 2 = (1)!/2 = 1/2
n=3
n=5
.
.
total 46 Prime Number in less than 200

And
In Exception of Even Composite Number n=4

n=4,
(4-1)!/4
= (3)!/4
= (1×2×3)/4
= 6/4
= 3/2