Let $\mathbb N\_{n} = \{1,2,\cdots,n\}$.


Let $S$ be of cardinality $n$ where elements of $S$ are integers from $\mathbb N\_{n}$ and at least one element of $S$ is repeated (That is at least one integer from $\mathbb N\_{n}$ is skipped. One can easily find a set $S$ with the property that:
$\displaystyle \sum\_{j \in S}j^{i} = \displaystyle \sum\_{j \in \mathbb N\_{n}}j^{i}$
when $i = 1$. (Example: $n=4$, $S=\{1,1,4,4\}$ has sum $10$, the same as sum of first n consecutive integers)


How about for $i \ge 2$? It is not obvious that higher power sum sets exist due to constraint in the cardinality of $S$ and $\mathbb N\_{n}$. One cannot deny it either? Is there a easy way to tackle some sumset questions?


For $i=2$ it is related to quadratic forms and integer norms. In an integer coordinate system, how many ways can a given integer norm occur when the coordinates are bounded?



Subset higher power sum question (related to quadratic forms)
It's a little easier to state an answer if you let $N\_n=\lbrace0,1,\dots,n-1\rbrace$. 


Let $n=2^k$, let $S$ be the multiset of integers with an odd number of ones in binary, each such integer appearing with multiplicity 2. Then it works for all $i\lt k$. 


E.g., $k=3$, $S=\lbrace1,2,4,7\rbrace$, each taken twice, you get $1^i+2^i+4^i+7^i=0^i+3^i+5^i+6^i$ for $i=0,1,2$ (where, by convention, $0^0=1$). 


If you really need the range to start at 1, just add 1 to everything, take $S=\lbrace2,3,5,8\rbrace$. 


This has to do with the Tarry-Escott problem, q.v.


