Find the remainder when the number of positive divisors of the value $$(3^{2020}+3^{2021})(3^{2021}+3^{2022})(3^{2022}+3^{2023})(3^{2023}+3^{2024})$$ is divided by $1000$.

[i]Proposed by pog[/i]
1. Let \( x = 3^{2020} \). Then the given expression can be rewritten as:
   \[
   (3^{2020} + 3^{2021})(3^{2021} + 3^{2022})(3^{2022} + 3^{2023})(3^{2023} + 3^{2024})
   \]
   Substituting \( x \) into the expression, we get:
   \[
   (x + 3x)(3x + 9x)(9x + 27x)(27x + 81x)
   \]
   Simplifying each term inside the parentheses:
   \[
   (4x)(12x)(36x)(108x)
   \]
   Combining these terms, we have:
   \[
   4 \cdot 12 \cdot 36 \cdot 108 \cdot x^4 = 4 \cdot 12 \cdot 36 \cdot 108 \cdot (3^{2020})^4
   \]
   Simplifying further:
   \[
   4 \cdot 12 \cdot 36 \cdot 108 \cdot 3^{8080}
   \]

2. Next, we find the prime factorization of the constant term:
   \[
   4 = 2^2, \quad 12 = 2^2 \cdot 3, \quad 36 = 2^2 \cdot 3^2, \quad 108 = 2^2 \cdot 3^3
   \]
   Combining these, we get:
   \[
   4 \cdot 12 \cdot 36 \cdot 108 = (2^2) \cdot (2^2 \cdot 3) \cdot (2^2 \cdot 3^2) \cdot (2^2 \cdot 3^3)
   \]
   Simplifying the powers of 2 and 3:
   \[
   2^2 \cdot 2^2 \cdot 2^2 \cdot 2^2 = 2^8
   \]
   \[
   3 \cdot 3^2 \cdot 3^3 = 3^{1+2+3} = 3^6
   \]
   Therefore:
   \[
   4 \cdot 12 \cdot 36 \cdot 108 = 2^8 \cdot 3^6
   \]
   Including the \( 3^{8080} \) term:
   \[
   2^8 \cdot 3^6 \cdot 3^{8080} = 2^8 \cdot 3^{8086}
   \]

3. To find the number of positive divisors of \( 2^8 \cdot 3^{8086} \), we use the formula for the number of divisors:
   \[
   (e_1 + 1)(e_2 + 1) \cdots (e_k + 1)
   \]
   where \( e_i \) are the exponents in the prime factorization. Here, \( e_1 = 8 \) and \( e_2 = 8086 \):
   \[
   (8 + 1)(8086 + 1) = 9 \cdot 8087
   \]

4. We need to find the remainder when \( 9 \cdot 8087 \) is divided by 1000:
   \[
   9 \cdot 8087 \equiv 9 \cdot 87 \pmod{1000}
   \]
   Simplifying \( 9 \cdot 87 \):
   \[
   9 \cdot 87 = 783
   \]

The final answer is \( \boxed{783} \).