To find the number of possible pairs of natural numbers that satisfy the given criteria, we need to understand the relationship between the highest common factor (HCF) and the least common multiple (LCM) of two numbers.

The HCF is the largest number that divides both numbers evenly, while the LCM is the smallest multiple that both numbers divide evenly.

Given that the HCF is 10 and the LCM is 1000, we can deduce the following:

1. The HCF of two numbers is a divisor of both numbers.
- This means that both numbers must be divisible by 10.

2. The LCM of two numbers is divisible by both numbers.
- This means that 1000 must be divisible by both numbers.

From these two points, we can conclude that both numbers must be factors of 1000 and multiples of 10.

Now, let's find all the factors of 1000.

The prime factorization of 1000 is 2^3 * 5^3.

To find all the factors, we can consider the exponents of the prime factors:
- The exponent for 2 can take values from 0 to 3 (0, 1, 2, 3).
- The exponent for 5 can also take values from 0 to 3 (0, 1, 2, 3).

Using these exponents, we can form pairs of numbers that satisfy the given criteria.

The possible pairs are:
- (2^0 * 5^0 * 10, 2^3 * 5^3 / (2^0 * 5^0 * 10)) = (1, 200)
- (2^1 * 5^0 * 10, 2^3 * 5^3 / (2^1 * 5^0 * 10)) = (2, 100)
- (2^2 * 5^0 * 10, 2^3 * 5^3 / (2^2 * 5^0 * 10)) = (4, 50)
- (2^3 * 5^0 * 10, 2^3 * 5^3 / (2^3 * 5^0 * 10)) = (8, 25)

Therefore, there are 4 possible pairs of natural numbers satisfying the given criteria.

Hence, the correct answer is option 'A' (2).