The $\textit{arithmetic derivative}$ $D(n)$ of a positive integer $n$ is defined via the following rules:
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[*] $D(1) = 0$;

[*] $D(p)=1$ for all primes $p$;

[*] $D(ab)=D(a)b+aD(b)$ for all positive integers $a$ and $b$.

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Find the sum of all positive integers $n$ below $1000$ satisfying $D(n)=n$.
1. **Understanding the arithmetic derivative:**
   - The arithmetic derivative \( D(n) \) is defined by:
     - \( D(1) = 0 \)
     - \( D(p) = 1 \) for all primes \( p \)
     - \( D(ab) = D(a)b + aD(b) \) for all positive integers \( a \) and \( b \)

2. **Applying the product rule:**
   - For \( n = p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k} \), where \( p_i \) are primes and \( a_i \) are positive integers, we use the product rule:
     \[
     D(n) = D(p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k}) = \sum_{i=1}^{k} D(p_i^{a_i}) \prod_{j \neq i} p_j^{a_j}
     \]

3. **Using the power rule:**
   - For \( D(p_i^{a_i}) \), we use the power rule:
     \[
     D(p_i^{a_i}) = a_i p_i^{a_i - 1} D(p_i) = a_i p_i^{a_i - 1} \cdot 1 = a_i p_i^{a_i - 1}
     \]

4. **Combining the rules:**
   - Substituting \( D(p_i^{a_i}) \) into the product rule:
     \[
     D(n) = \sum_{i=1}^{k} a_i p_i^{a_i - 1} \prod_{j \neq i} p_j^{a_j}
     \]
   - Simplifying, we get:
     \[
     D(n) = n \sum_{i=1}^{k} \frac{a_i}{p_i}
     \]

5. **Setting \( D(n) = n \):**
   - We need \( D(n) = n \), so:
     \[
     n \sum_{i=1}^{k} \frac{a_i}{p_i} = n \implies \sum_{i=1}^{k} \frac{a_i}{p_i} = 1
     \]

6. **Analyzing the equation:**
   - For the sum \( \sum_{i=1}^{k} \frac{a_i}{p_i} = 1 \) to hold, each term \( \frac{a_i}{p_i} \) must be an integer. This implies \( a_i = p_i \) and all other \( a_j = 0 \) for \( j \neq i \).

7. **Finding valid \( n \):**
   - Therefore, \( n \) must be of the form \( p^p \) where \( p \) is a prime.
   - We need to find all such \( n < 1000 \).

8. **Calculating \( p^p \) for primes \( p \):**
   - For \( p = 2 \):
     \[
     2^2 = 4
     \]
   - For \( p = 3 \):
     \[
     3^3 = 27
     \]
   - For \( p = 5 \):
     \[
     5^5 = 3125 \quad (\text{which is greater than 1000})
     \]

9. **Summing valid \( n \):**
   - The valid \( n \) are \( 4 \) and \( 27 \).
   - Their sum is:
     \[
     4 + 27 = 31
     \]

The final answer is \(\boxed{31}\).