Because none of the six consecutive positive integers is divisible with $7$ they are of this kind: $7n+1$, $7n+2$, $7n+3$, $7n+4$, $7n+5$, $7n+6$, $n \in \mathbb{N}_0$. Their sum is $S=42n+21=21(2n+1)$, from where it follows that $S$ is divisible with $21$ but it's not divisible with $42$. In order $S$ to be a square of a positive integer it must $2n+1=21k^2$ for some odd number $k$ and in order $S$ to be a four-digit number the inequality $2<k^2<23$ must hold. This is possible only if $k^2=9$ from where we obtain that $2n+1=21k^2=189$ and $n=94$. The desired numbers are $659$, $660$, $661$, $662$, $663$ and $664$.