We will prove that the numbers congruent to $0,4$ or $8$ in mod $9$ satisfy the condition. (I don't have an example for these numbers, but they satisfy the condition) First of all, let's prove that $\dfrac{x_1^2+x_2^2+\cdots+x_n^2}{(x_1+2x_2+\cdots+nx_n)^2}\leq \dfrac{27}{4n(n+1)(2n+1)}$. We know that $1^2+2^2+\cdots+n^2=\dfrac{n(n+1)(2n+1)}{6}\Rightarrow \dfrac{27}{4n(n+1)(2n+1)}=\dfrac{9}{8(1^2+2^2+\cdots+n^2)}$. Then, we should prove that $\dfrac{x_1^2+x_2^2+\cdots+x_n^2}{(x_1+2x_2+\cdots+nx_n)^2}\leq \dfrac{9}{8(1^2+2^2+\cdots+n^2)}\Leftrightarrow$ $8(x_1^2+x_2^2+\cdots+x_n^2)(1^2+2^2+\cdots+n^2)\leq 9(x_1+2x_2+\cdots+nx_n)^2=(3x_1+6x_2+\cdots+3nx_n)^2$. We have $i\leq x_i\leq 2i, \forall i\in \{1,2,\cdots,n\}$. Then $(x_i-i)(2i-x_i)\ge 0\Rightarrow 3ix_i\ge x_i^2+2i^2$. Then, $(3x_1+6x_2+\cdots+3nx_n)^2\ge (x_1^2+x_2^2+\cdots+x_n^2+2(1^2+2^2+\cdots+n^2))^2$. Let $A=x_1^2+x_2^2+\cdots+x_n^2$ and $B=1^2+2^2+\cdots+n^2$. It suffices to prove that $8AB\leq (A+2B)^2$. But, $8AB\leq (A+2B)^2\Leftrightarrow (A-2B)^2\ge 0$, which is correct. This completes the proof. We should have equality. Then $x_i=i\lor x_i=2i, \forall i\in \{1,2,\cdots,n\}$ and $x_1^2+x_2^2+\cdots+x_n^2=2(1^2+2^2+\cdots+n^2)$. $x_i=i\lor x_i=2i\Rightarrow (x_i-i)(2i-x_i)=0\Rightarrow 3ix_i=x_i^2+2i^2$. Then, $x_1^2+x_2^2+\cdots+x_n^2=3(x_1+2x_2+\cdots+nx_n)-2(1^2+2^2+\cdots+n^2)$. Hence, $3(x_1+2x_2+\cdots+nx_n)-2(1^2+2^2+\cdots+n^2)=2(1^2+2^2+\cdots+n^2)\Rightarrow$ $3(x_1+2x_2+\cdots+nx_n)=4(1^2+2^2+\cdots+n^2)\Rightarrow 3|4(1^2+2^2+\cdots+n^2)\Rightarrow 3|1^2+2^2+\cdots+n^2$. We have $1^2+2^2+\cdots+n^2=\dfrac{n(n+1)(2n+1)}{6}$. Then, $9|n(n+1)(2n+1)$. If $n\equiv 0\pmod{3}$, then $9|n$. Hence, $n\equiv 0\pmod{9}$ If $n\equiv 1\pmod{3}$, then $9|2n+1$. Hence, $n\equiv 4\pmod{9}$ If $n\equiv 2\pmod{3}$, then $9|n+1$. Hence, $n\equiv 8\pmod{9}$ Thus, we conclude that $n\equiv 0,4,8\pmod{9}$.

any solution?

To prove that the final integer function have solution, i think we can use generating function. BarisKoyuncu wrote: $3(x_1+2x_2+\cdots+nx_n)=4(1^2+2^2+\cdots+n^2)

laikhanhhoang_3011 wrote: To prove that the final integer function have solution, i think we can use generating function. BarisKoyuncu wrote: $3(x_1+2x_2+\cdots+nx_n)=4(1^2+2^2+\cdots+n^2)$

Jjesus wrote: any solution? hint: for every 27 consecutive integers set $S$, there exists $A,B$ which is a partition of $S$,s.t.$|A|=9,|B|=18$, and $2\sum_{x\in A}x^2=\sum_{x\in B}x^2$. For small cases, I think it's easy.

Here is the constructions for $n\equiv 0, 4, 8 \pmod{9}$ by induction. $S$ will contain the $i$s such that $x_i=2i$ $$n=4 \Rightarrow S=\{1, 3 \}$$$$n=8 \Rightarrow S=\{2, 8 \}$$$$n=9 \Rightarrow S=\{1, 2, 3, 9 \}$$$$n=13 \Rightarrow S=\{2, 10, 13 \}$$$$n=17 \Rightarrow S=\{1, 7, 16, 17\}$$$$n=18 \Rightarrow S=\{3, 9, 17, 18 \}$$satisfies. For $n=k$, assume that we have found suitable sets and let's try to find $n=k+18$. We will have $$\frac{(k+18)(k+19)(2k+37)}{18}- \frac{k(k+1)(2k+1)}{18}= (k+1)^2+(k+4)^2+(k+11)^2+(k+12)^2+(k+14)^2+(k+15)^2$$Hence $S'=S \cup \{k+1, k+4, k+11, k+12, k+14, k+15 \}$ will satisfy for $n=k+18$.

electrovector very nice construction

i think this expression is usually hard to come by in the brain, so it must be p6.