A 6-digit number composed of consecutive natural numbers can be expressed as \( n, n+1, n+2, n+3, n+4, n+5 \). The sum of these digits is calculated as follows:
\[ S = n + (n+1) + (n+2) + (n+3) + (n+4) + (n+5) \]
\[ S = 6n + 15 \]
The expression \( 6n + 15 \) is invariably divisible by 3 due to the following reasons:
- The product of any integer \( n \) and 6 is divisible by 3, since 6 is a multiple of 3.
- The constant value 15 is itself divisible by 3.
Consequently, \( 6n + 15 \) is divisible by 3, which in turn means the original 6-digit number is always divisible by 3.