The sum of the (decimal) digits of a natural number $n$ equals $100$, and the sum of digits of $44n$ equals $800$.  Determine the sum of digits of $3n$.
1. Let \( n \) be a natural number such that the sum of its digits is 100. We can represent \( n \) in its decimal form as:
   \[
   n = 10^k a_k + 10^{k-1} a_{k-1} + \ldots + 10 a_1 + a_0
   \]
   where \( a_i \) are the digits of \( n \). Therefore, the sum of the digits of \( n \) is:
   \[
   a_k + a_{k-1} + \ldots + a_1 + a_0 = 100
   \]

2. Given that the sum of the digits of \( 44n \) equals 800, we need to analyze the representation of \( 44n \). We can write:
   \[
   44n = 44 \cdot (10^k a_k + 10^{k-1} a_{k-1} + \ldots + 10 a_1 + a_0)
   \]
   Expanding this, we get:
   \[
   44n = 10^{k+1} (4a_k) + 10^k (4a_k + 4a_{k-1}) + 10^{k-1} (4a_{k-1} + 4a_{k-2}) + \ldots + 10 (4a_1 + 4a_0) + 4a_0
   \]

3. The sum of the digits in this representation is:
   \[
   4a_k + 4a_k + 4a_{k-1} + 4a_{k-1} + \ldots + 4a_0 + 4a_0 = 800
   \]
   Simplifying, we get:
   \[
   8(a_k + a_{k-1} + \ldots + a_1 + a_0) = 800
   \]
   Since \( a_k + a_{k-1} + \ldots + a_1 + a_0 = 100 \), we have:
   \[
   8 \cdot 100 = 800
   \]

4. To ensure there are no carries in the sum of the digits of \( 44n \), we must have \( 4a_i + 4a_{i-1} < 10 \) for all \( i \). This implies:
   \[
   a_i + a_{i+1} \leq 2
   \]
   for all \( i \).

5. Now, we need to determine the sum of the digits of \( 3n \). Since \( a_i + a_{i+1} \leq 2 \), multiplying by 3 will not cause any carries. Therefore, the sum of the digits of \( 3n \) is:
   \[
   3 \cdot (a_k + a_{k-1} + \ldots + a_1 + a_0) = 3 \cdot 100 = 300
   \]

The final answer is \(\boxed{300}\).