Let $a_i$ be the $i$-th symmetric sum and let $b_i=a^i+b^i+c^i+d^i.$ Consider $$P(x)=x^4-a_1x^3+a_2x^2-a_3x+a_4$$such that $a,b,c,d$ are roots of $P.$ Then the Newton tells us: $$b_4-a_1b_3+a_2b_2-a_3b_1+4a_4=0$$Which means that it suffices to show that $n|-a_1b_3+a_2b_2-a_3b_1.$ But by the condition itself we have $n|b_1,n|b_2,n|a_1$ and thus $$n|b_4+4a_4= a^4 + b^4 + c^4 + d^4 + 4abcd$$so we are done. $\blacksquare$

Very easy.From Newton's identity,we have $$s_4-s_3e_1+s_2e_2-s_1e_3+4e_4=0$$ consider in modulo $n$,it's obvious that $LHS\equiv s_4+4e_4\text{(mod n)}$ and $s_4+4e_4\equiv 0\text{(mod n)}$

\[ \begin{split} &(a+b+c+d)(-a^2b-a^2c-a^2d-b^2a-b^2c-b^2d-c^2a-c^2b-c^2d-d^2a-d^2b-d^2c\\ &+abc+abd+acd+bcd)+(a^2+b^2+c^2+d^2)(a^2+b^2+c^2+d^2+ab+ac+ad+bc+bd+cd)\\ =&a^4+b^4+c^4+d^4+4abcd \end{split} \] So we have done