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 $x_1^2 + x_2^2 + x_3^2 + x_4^2 \equiv 7 \pmod{8}$ 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.