Let us call a natural number rare if $n$ satisfies the given condition. Clearly $n=1$ and $n=2$ are rare numbers. Next assume $n>2$ is a rare number. Then there are $n-1$ primes $7 < p_1 < p_2 < \ldots \ p_{n-1}$ s.t. $n \mid 5^2 + \sum_{i=1}^{n-1} p_i^2$ and $n \mid 7^2 + \sum_{i=1}^{n-1} p_i^2$, yielding $n \mid 7^2 - 5^2 = 49 - 25$, i.e. $(1) \;\; n \mid 24$. Assume $q_1,q_2, \ldots ,q_n$ are $n$ distinct primes greater than 3. Then for $m \in \{3,8\}$ $S = \sum_{i=1}^n q_i^2 \equiv \sum_{i=1}^n 1 = n \pmod{m}$, i.e. $(2) \;\; m \mid S - n$. The fact that 3 and 8 are coprime combined with (2) result in $(3) \;\; 24 \mid S - n$. We know that $n | S$ (since $n$ is a rare), implying $n | S-n$. Hence by (1) and (3) every $n$ satisfying (1) is a rare number. Conclusion The rare numbers are the positive divisors of 24, i.e. $1,2,3,4,6,8,12,24$.