Let those $8$ consecutive positive integers be $n,n+1,n+2,n+3,n+4,n+5,n+6,n+7$. Let $x_i,i={1,2,3,4,5,6,7,8}$ be a permutation of the above numbers for which holds that $x_1^2+x_2^2+x_3^2+x_4^2=x_5^2+x_6^2+x_7^2+x_8^2$. We also know that $\sum_{i=1}^{8}x_i^2=8n^2+56n+140$, so that $x_1^2+x_2^2+x_3^2+x_4^2=x_5^2+x_6^2+x_7^2+x_8^2=4n^2+28n+70$. Let us look at the power of two of the numbers $n,n+1,n+2,n+3,n+4,n+5,n+6,n+7$: $n^2=n^2+0n+0$ $(n+1)^2=n^2+2n+1$ $(n+2)^2=n^2+4n+4$ $(n+3)^2=n^2+6n+9$ $(n+4)^2=n^2+8n+16$ $(n+5)^2=n^2+10n+25$ $(n+6)^2=n^2+12n+36$ $(n+7)^2=n^2+14n+49$ The sum of the constans of the those numbers have to be equal on both sides to 70, so lets look which numbers can be with $(n+7)^2$ on the same side. Beacuse $70-49=21$ we can eliminate $n+6$ and $n+5$. Now let's look at the sums of the numbers ${0,1,4,9,16}$ taking only 3 elements at a time: $0+1+4=5$ $0+1+9=10$ $0+1+16=17$ $0+4+9=13$ $0+4+16=20$ $0+9+16=25$ $1+4+9=14$ $1+4+16=21$ $1+9+16=26$ $4+9+16=29$ We clearly see that only $(n+1)^2,(n+2)^2,(n+4)^2$ satesfy the condition to be on the same side with $(n+7)^2$ so we have now: $(n+1)^2+(n+2)^2+(n+4)^2+(n+7)^2=n^2+(n+3)^2+(n+5)^2+(n+6)^2=4n^2+28n+70$ Now we can clearly see that the numbers $n+1,n+2,n+4,n+7$ have to be in one set and the numbers $n,n+3,n+5,n+6$ in the other one. Now we have $(n+1)+(n+2)+(n+4)+(n+7)=4n+14=(n)+(n+3)+(n+5)+(n+6)$, Q.E.D.

I don't think the sum of the linear parts need to be 70 because the expression may be true only for a particular n.

gobathegreat wrote: Let $X$ be a set which consists from $8$ consecutive positive integers. Set $X$ is divided on two disjoint subsets $A$ and $B$ with equal number of elements. If sum of squares of elements from set $A$ is equal to sum of squares of elements from set $B$, prove that sum of elements of set $A$ is equal to sum of elements of set $B$. It is easy to show that the only four-elements subset $A\subset\{0,1,2,3,4,5,6,7\}$ such that $\sum_{x\in A}x>14$ and $\sum_{x\in A}x^2< 70$ is $A=\{2,3,4,6\}$ Suppose now that $\exists n>0$ and $\{u_k\}_{k=1}^8$ permutation of $(0,1,2,3,4,5,6,7)$ such that : $\sum_{i=1}^4(n+u_i)^2=\sum_{i=5}^8(n+u_i)^2$ while $\sum_{i=1}^4(n+u_i)\ne\sum_{i=5}^8(n+u_i)$ WLOG $\sum_{i=1}^4(n+u_i)>\sum_{i=5}^8(n+u_i)$ This would mean : $\sum_{i=1}^4u_i>\sum_{i=5}^8u_i=28-\sum_{i=1}^4u_i$ and so $\sum_{i=1}^4u_i>14$ $2n\sum_{i=1}^4u_i+\sum_{i=1}^4u_i^2=2n\sum_{i=5}^8u_i+\sum_{i=5}^8u_i^2$ and so $\sum_{i=1}^4u_i^2<\sum_{i=5}^8u_i^2=140-\sum_{i=1}^4u_i^2$ and so $\sum_{i=1}^4u_i^2<70$ And so $(u_1,u_2,u_3,u_4)=(2,3,4,6)$ and $(u_5,u_6,u_7,u_8)=(0,1,5,7)$ But then $\sum_{i=1}^4(n+u_i)^2=\sum_{i=5}^8(n+u_i)^2$ becomes $30n+65=26n+75$ and so no such $n$ Q.E.D.