To solve the problem, we need to find positive integers (n) such that (n) is a multiple of 1000 ((2^3 \times 5^3)) and the number of divrs (d(n)) is a multiple of 1000 ((2^3 \times 5^3)).

Key Observations:

• For (n = 2^a 3^b 5^c ...), (a \geq 3) and (c \geq 3) (since 1000 divides (n)).
• (d(n) = (a+1)(b+1)(c+1)...) must be divisible by (2^3 \times 5^3).

Minimal (n) Calculation:

We need (d(n)) to have at least three factors of 2 and three factors of 5. The optimal way to minimize (n) is to assign exponents to the smallest primes first.

The product ((a+1)(b+1)(c+1)...) needs to include (2^3) and (5^3). The most efficient combination is:

• ((a+1)=5) ((a=4)), ((b+1)=5) ((b=4)), ((c+1)=5) ((c=4)) (contributing (5^3)), and ((d+1)=8) ((d=7)) (contributing (2^3)).

Thus, (n = 2^4 \times 3^4 \times 5^4 \times 7^7).

Computation of Minimal (n):

[2^4 = 16, \quad 3^4=81, \quad 5^4=625, \quad 7^7=823543]
[n = 16 \times 81 \times 625 \times 823543 = 667069830000]

Answer: (\boxed{667069830000}) (assuming the question asks for the minimal such (n), given the ambiguity in the original problem statement).

(\boxed{667069830000})

But wait, if the question indeed asks for the count of such (n) (with a missing constraint), perhaps the intended answer is different. However, given the context, the minimal (n) is the most plausible answer here.

Final Answer: (\boxed{667069830000})

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