Define $S_m(n)=1^m+2^m+\cdots+n^m$
Can it be shown that
$S_{2m}(uv)\equiv0\pmod{uv}\iff S_{2m}(u)\equiv0\pmod{u}$ and $S_{2m}(v)\equiv0\pmod{v}$
Where $m,u,v$ are positive integer
Example:$1|S_2(1)=1,5|S_2(5)=55,7|S_2(7)=140,25|S_2(25)=5525,35|S_2(35)=14910,49|S_2(49)=40425$.
Note: if $p$ prime, $p\nmid S_{2m}(p)$ then $p-1\mid 2m$ Proof
I hope following my another claim helps to solve this
Consider $n\equiv 1\pmod2$ then $S_{2m}(n)\equiv0\pmod{n}\iff S_{2m}(\frac{n-1}2)\equiv0\pmod{n}$
Formula: $$S_m(n)=\sum_{k=1}^{n} k^{m}=\sum_{b=1}^{m+1} \binom{n}b\sum_{i=0}^{b-1} (-1)^{i}(b-i)^{m}\binom{b-1}i$$
Source code Pari/GP
for(m=1,30,for(a=2,100,if(sum(i=1,a,i^(2*m))%a==0,print([2*m,a,sum(i=1,a,i^(2*m))]))))
The problem is posted in MO link
