Team Selection Test

The answer is all positive integers except $1$, $2$ and $4$.

Let us call a positive integer good if it satisfies the condition given in the problem. We first show that if $n$ is good, so is any multiple of $n$.

Let $m=nk$ and $x_1,x_2,\dots ,x_m$ be integers such that $m\mid x_i$ for all $1\le i\le m$. Then since $n\mid x_i$ for all $1\le i\le m$ and $n$ is good, there exist integers $y_1,y_2,\dots ,y_m$ such that $$\sum _{i=nl+1}^{n(l+1)}{x}_i^2=\sum _{i=nl+1}^{n(l+1)}{y}_i^2$$ for all $0\le l\le k-1$ and $n\nmid y_i$ for all $1\le i\le m$. Therefore we obtain that $$\sum _{i=1}^{m=nk}{x}_i^2=\sum _{i=1}^{m=nk}{y}_i^2$$ and $m=nk\nmid y_i$ for all $1\le i\le m$.

Lemma: Let $n$ be a positive odd integer and $x_1,x_2,\dots ,x_n$ be integers with at least one of them is not divisible by $n$. Then there exist integers $y_1,y_2,\dots ,y_n$ such that none of them is divisible by $n$ and $$\sum _{i=1}^n(n{x}_i{)}^2=\sum _{i=1}^n{y}_i^2.$$ Proof: Without loss of generality we may assume that $n\nmid x_1$. Let $X=2{\sum }_{i=1}^n{x}_i$. If $n\nmid X$, then replace $x_1$ by $-x_1$. As $n\nmid x_1$ and $n$ is odd, $n\nmid 4x_1$ and hence we may assume that $n\nmid X$. Then by the following identity $$\sum _{i=1}^n(n{x}_i{)}^2=\sum _{i=1}^n(X-n{x}_i{)}^2$$ letting $y_i=X-n{x}_i$ for all $1\le i\le n$ works.

For a positive odd integer $n$, if a positive integer $a$ is sum of squares of $n$ integers with each of them is divisible by $n$, then there exist integers $x_1,x_2,\dots ,x_n$ and a positive integer $r$ such that $a={\sum }_{i=1}^n(n^r{x}_i{)}^2$ and $n\nmid x_i$ for some $1\le i\le n$. Applying the lemma $r$ times we can find integers $y_1,y_2,\dots ,y_n$ such that $a={\sum }_{i=1}^n{y}_i^2$ and $n\nmid y_i$ for all $1\le i\le n$.

Next we show that $8$ is good. Let $a$ be positive integer which is sum of squares of $8$ integers with each of them is divisible by $8$. Then $64\mid a$, hence $a\ge 64$ and $a=1^2+4^2+4^2+4^2+x_1^2+x_2^2+x_3^2+x_4^2$ for some integers $x_1,x_2,x_3,x_4$ by Lagrange's four-square theorem. Note that $x_1^2+x_2^2+x_3^2+x_4^2\equiv 7(\mathrm{mod}8)$ and the only way to get $7$ as sum of four quadratic residues in (mod $8$) is $1+1+1+4$. Therefore, $8\nmid x_i$ for all $1\le i\le 4$.

Finally, we observe that $32=4^2+4^2+0^2+0^2$ is a counterexample for $4$ and we are done.