Goutham wrote: $(CZS 2)$ Let $p$ be a prime odd number. Is it possible to find $p-1$ natural numbers $n + 1, n + 2, . . . , n + p -1$ such that the sum of the squares of these numbers is divisible by the sum of these numbers? Write the numbers as $x-\frac{p-3}{2},x-\frac{p-5}{2},\cdots,x+\frac{p-1}{2}.$ Then we have to get $(p-1)x+\frac{p-1}{2}|\frac{p-1}{2}x+\frac{p-1}{2}\frac{p-3}{3}\frac{p-2}{2}+(\frac{p-1}{2})^2$ For $p=3$ this gives $2x+1|2x^2+2x+1$ but that is simple not possible. For $p\equiv 5 \pmod{6}$ $x=\frac{p-3}{3}\frac{p-2}{2}+\frac{p-1}{2}$ is a solution. For $p\equiv 1 \pmod{6}$ it has clearly no solutions, because RL is not a multiple of $\frac{p-1}{2}.$ So ansdwer: only if $p\equiv 5 \pmod{6}$ ex. $p=5$ gives $1+2+3+4|30$