Introduction:
Arithmetic Progression is a sequence of numbers in which each term after the first is obtained by adding a constant to the preceding term. In this problem, we need to find out the number of natural numbers between 200 and 500 which are divisible by 7.

Method:
We can solve this problem by using the formula for the nth term of an arithmetic progression, which is given by:

An = A1 + (n-1)d

Where An is the nth term, A1 is the first term, n is the number of terms, and d is the common difference between the terms.

We know that the first term in this case is 203 (the smallest multiple of 7 greater than 200) and the common difference is 7 (since we are looking for multiples of 7). We can find the number of terms by using the formula:

n = (An - A1)/d + 1

Where An is the largest multiple of 7 less than or equal to 500.

Solution:
Using the above formulas, we get:

An = 497
n = (497 - 203)/7 + 1 = 42

Therefore, there are 42 natural numbers between 200 and 500 which are divisible by 7. They are:

203, 210, 217, 224, 231, 238, 245, 252, 259, 266, 273, 280, 287, 294, 301, 308, 315, 322, 329, 336, 343, 350, 357, 364, 371, 378, 385, 392, 399, 406, 413, 420, 427, 434, 441, 448, 455, 462, 469, 476, 483, 490, 497.

Therefore, the answer is 42.