1. The sum of six consecutive natural numbers, none of which is divisible by 7, is divisible by 21, but not by 42. Prove it!

Determine six such numbers, such that their sum is a four-digit number and is the square of some natural number.
The numbers are $7n+1, 7n+2, 7n+3, 7n+4, 7n+5, 7n+6$ and their sum is $S=42n+21=21(2n+1)$. Clearly, $21 \mid S$ and when $S$ is divided by 42, the remainder is 21, i.e., $S$ is not divisible by 42.

For this sum to be a square of a natural number, we need $2n+1=21k^2$, where $k$ is an odd natural number. For this sum to be a four-digit number, we must have $2 < k^2 < 23$. This is possible only for $k^2=9$. Now we get $2n+1=21k^2=189$, i.e., $n=94$. The required numbers are: $659, 660, 661, 662, 663$ and 664.