Answer. Any square-free number.

Since $S_j$ doesn't exist for $j \le 1$, the statement is held for $i = 1$. On the other hand if for an integer $n > 1$ there exists a prime number $p$, which $p^2 \mid n$, $n$ is not a valid number; Because for all integer numbers $x, y$
$$x^2 + n y^2 = x^2 + \left(\frac{n}{p^2}\right) (y p)^2.$$
which means that
$$S_n \subseteq S_{\frac{n}{p^2}}.$$
Now we prove that any square-free number has the property.

Lemma. If $n$ is a square-free positive integer then there are distinct odd prime numbers $p_k$ for $1 \le k \le n-1$ where $-n$ is a quadratic residue modulo $p_k$ but $-k$ is not. i.e. in Legendre symbol
$$\left(\frac{-n}{p_k}\right) = 1, \left(\frac{-k}{p_k}\right) = -1. \qquad (1)$$
Proof. Assume that $n = q_1 \dots q_m$, where $q_i$'s are prime numbers. The claim is that at least one of the $q_i$'s does not divide $k$; otherwise if for $1 \le i \le m$ $q_i$ divides $k$ then
$$n \mid k \longrightarrow n \le k;$$
Which is a contradiction.
Without loss of generality assume that $q_m$ is one of the numbers that does not divide $k$. So $k = q_1 \dots q_r \ell_1 \dots \ell_t D^2$ where $0 \le r < m$.
If $q_1 \dots q_m \ell_1 \dots \ell_t$ is an odd number then the numbers $4, q_1, \dots, q_m, \ell_1, \dots, \ell_t$ are pairwise coprime so by applying the Chinese remainder theorem there exists an integer $x$ where
$$\begin{cases} x \equiv 3 \pmod{4} \\ x \equiv -1 \pmod{q_i} & 1 \le i \le m-1 \\ x \equiv u \pmod{q_m} \\ x \equiv -1 \pmod{\ell_i} & 1 \le i \le t. \end{cases}$$
Where $u$ is an integer number which $-u$ is a non-quadratic residue modulo $q_m$. Notice that thus $q_m$ is an odd prime number such a number exists.
Because $x$ and $4q_1 \dots q_m \ell_1 \dots \ell_t$ are coprime, By applying Dirichlet theorem,
there are infinitely many prime numbers of the form $x + 4q_1 \dots q_m \ell_1 \dots \ell_t d$.
Take one of those like $p_k$ where $p_k > \max(n, 3)$.
By applying the law of quadratic reciprocity for odd prime numbers $p_k$ and $q_j$
$$\left(\frac{p_k}{q_j}\right) \left(\frac{q_j}{p_k}\right) = (-1)^{\frac{(p_k-1)(q_j-1)}{4}}. \qquad (2)$$
$$p_i \equiv 3 \pmod{4} \implies \frac{p_k - 1}{2} \equiv 1 \pmod{2}$$
So
$$(-1)^{\frac{(p_k-1)(q_j-1)}{2}} = (-1)^{\frac{q_j-1}{2}}.$$
It's easy to check that
$$\left(\frac{-1}{q_j}\right) = (-1)^{\frac{q_j-1}{2}}.$$
Also for $1 \le j \le m-1$
$$\left(\frac{p_k}{q_j}\right) = \left(\frac{-1}{q_j}\right).$$
Then according to (2)
$$\left(\frac{p_k}{q_j}\right) \left(\frac{q_j}{p_k}\right) = \left(\frac{-1}{q_j}\right) \left(\frac{q_j}{p_k}\right) = (-1)^{\frac{(p_k-1)(q_j-1)}{4}} = (-1)^{\frac{q_j-1}{2}} = \left(\frac{-1}{q_j}\right).$$
So
$$\left(\frac{q_j}{p_k}\right) = 1 \quad \forall j: 1 \le j \le m-1$$
In the same way
$$\left(\frac{\ell_s}{p_k}\right) = 1 \quad \forall s: 1 \le s \le t.$$
In addition
$$\left(\frac{p_k}{q_m}\right) \left(\frac{q_m}{p_k}\right) = (-1)^{\frac{(p_k-1)(q_m-1)}{4}} = (-1)^{\frac{q_m-1}{2}} = \left(\frac{-1}{q_m}\right).$$
Also
$$\left(\frac{p_k}{q_m}\right) \left(\frac{q_m}{p_k}\right) = \left(\frac{u}{q_m}\right) \left(\frac{q_m}{p_k}\right) = \left(\frac{-1 \times -u}{q_m}\right) \left(\frac{q_m}{p_k}\right) = -\left(\frac{-1}{q_m}\right) \left(\frac{q_m}{p_k}\right)$$
So
$$-\left(\frac{-1}{q_m}\right) \left(\frac{q_m}{p_k}\right) = \left(\frac{-1}{q_m}\right) \longrightarrow \left(\frac{q_m}{p_k}\right) = -1$$
Since the Legendre symbol is a completely multiplicative function of its top argument, then
$$\begin{aligned} \left(\frac{-k}{p_k}\right) &= \left(\frac{-1}{p_k}\right) \prod_{j=1}^r \left(\frac{q_j}{p_k}\right) \prod_{s=1}^t \left(\frac{\ell_s}{p_k}\right) \left(\frac{D^2}{p_k}\right) = \left(\frac{-1}{p_k}\right) \prod_{j=1}^r \prod_{s=1}^t 1 = -1, \\ \left(\frac{-n}{p_k}\right) &= \left(\frac{-1}{p_k}\right) \prod_{j=1}^m \left(\frac{q_j}{p_k}\right) = \left(\frac{-1}{p_k}\right) \left(\frac{q_m}{p_k}\right) \prod_{j=1}^{m-1} 1 = -1 \times -1 = 1. \end{aligned}$$
So if $n$ and $\ell_1 \dots \ell_t$ are odd numbers, $p_k$ has been found. If one is $\ell_j$'s is two; Assume that $\ell_t = 2$ and take
$$\left\{ \begin{array}{ll} x \equiv 7 \pmod{8} \\ x \equiv -1 \pmod{q_i} & 1 \le i \le m-1 \\ x \equiv u \pmod{q_m} \\ x \equiv -1 \pmod{\ell_i} & 1 \le i \le t-1. \end{array} \right.$$
Since $q_m$ does not divide $k$ then it is an odd prime number and like before $u$ exists. Because $x$ and $8q_1 \dots q_m \ell_1 \dots \ell_{t-1}$ are coprime, By applying Dirichlet theorem, there are infinitely many prime numbers of the form $x + 8q_1 \dots q_m \ell_1 \dots \ell_{t-1}d$. Take one of those like $p_k$ where $p_k > \max(n, 3)$. Thus
$$\left(\frac{2}{p_k}\right) = (-1)^{\frac{p_k^2-1}{8}} = 1$$
So
$$\left(\frac{\ell_1}{p_k}\right) = 1$$
and the proof is the same as before.
Also if one of the $q_1, q_2, \dots, q_{m-1}$ is two the proof is still as before.
Now assume that $q_m = 2$.
$$\left\{ \begin{array}{ll} x \equiv 3 \pmod{8} \\ x \equiv -1 \pmod{q_i} & 1 \le i \le m-1 \\ x \equiv -1 \pmod{\ell_i} & 1 \le i \le t-1. \end{array} \right.$$
In this case (By applying Dirichlet theorem we can take $p_k$ similarly)
$$\left(\frac{2}{p_k}\right) = (-1)^{\frac{p_k^2-1}{8}} = -1.$$
So
$$\left(\frac{q_m}{p_k}\right) = -1.$$
And the proof of the lemma is done.

Now back to the main problem. Take $p_1, p_2, \dots, p_{n-1}$ as lemma says.
Because $-n$ is a quadratic residue mod $p_i$, there exists an integer number $a$
such that
$$a^2 \equiv -n \pmod{p_i} \implies a^2 + n \equiv 0 \pmod{p_i}.$$
Also
$$(a+p_i)^2 + n - (a^2+n) = p_i(2a+p_i).$$
Since $n \neq 0 \pmod{p_i}$ and $p_i$ is odd then $2a$ is not divisible by $p_i$ and then $2a+p_i$ is not divisible by $p_i$. So if $a^2+n$ is divisible by $p_i^2$ then $(a+p_i)^2+n$ is not and
$$(a+p_i)^2+n \equiv 0 \pmod{p_i}.$$
So there exists an integer number $x_i \in \{a, a+p\}$ which
$$\begin{cases} (x_i)^2 + n \equiv 0 \pmod{p_i} \\ (x_i)^2 + n \not\equiv 0 \pmod{p_i^2}. \end{cases}$$
The numbers $p_1^2, p_2^2, \dots, p_{n-1}^2$ are pairwise coprime so by applying Chinese remainder theorem there exists an integer number $x$ such that
$$x \equiv x_k \pmod{p_k^2} \quad \forall j: 2 \le j \le l.$$
Claim 1. The number $x^2+n$ doesn't belong to any of the subsets $S_1, S_2, \dots, S_n$.
Proof. If not, then there is an integer $1 \le r \le n-1$ which $x^2+n \in S_r$. Because $1 \le r \le n-1$ so take $p_r$. Also there exists integer numbers $u, v$ such that
$$x^2 + n = u^2 + r v^2. \qquad (3)$$
By the definition of $p_r$
$$\begin{cases} x \equiv x_r \pmod{p_r}, \\ (x_r)^2 + n \equiv 0 \pmod{p_r}, \\ (x_r)^2 + n \not\equiv 0 \pmod{p_r^2}. \end{cases} \qquad (4)$$
According to equation (3)
$$u^2 + r v^2 \equiv 0 \pmod{p_r}.$$
Since $r \neq 0 \pmod{p_r}$, then
$$u \equiv 0 \pmod{p_r} \iff v \equiv 0 \pmod{p_r}.$$
So if one of them is divisible by $p_i$ then the another one is too and then the whole number $u^2 + r v^2$ will be divisible by $p_i^2$ which contradicts with (4). Also
$$u^2 + r v^2 \equiv 0 \pmod{p_r} \implies u^2 \equiv (-r)v^2 \pmod{p_r}.$$
By Taking $\frac{p_r-1}{2}$ exponent of each side of the last equivalence
$$u^{p_r-1} \equiv (-r)^{\frac{p_r-1}{2}} v^{p_r-1} \pmod{p_r}. \qquad (5)$$
Since $u, v \not\equiv 0 \pmod{p_r}$ by Fermat's little theorem
$$u^{p_r-1} \equiv v^{p_r-1} \equiv 1 \pmod{p_r}$$
And from (5)
$$(-r)^{\frac{p_r-1}{2}} \equiv 1 \pmod{p_r}.$$
Let $g$ be the primitive root of the $\mathbb{Z}_p^*$ so there exists an integer number $m$ which
$$g^m \equiv -r \pmod{p_r}.$$
By Taking $\frac{p_r-1}{2}$ exponent of each side of the equivalence
$$g^{\frac{p_r-1}{2}} \equiv (-r)^{\frac{p_r-1}{2}} \equiv 1 \pmod{p_r}.$$
Since the order of $g$ is $p_r - 1$ in $\mathbb{Z}_p^*$
$$p_r - 1 \mid m \frac{p_r - 1}{2} \implies 2 \mid m.$$
So $m$ is an even number and then $-r$ is a quadratic residue which is a contradiction. This contradiction shows that our first assumption that $x^2 + n$ belongs to $S_r$ is wrong so $n$ has the property that is mentioned in the question. So all of the numbers with the property have been found. $\square$