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The problem is to specify all such $n>1$ that for any $a,b,c,d\in\mathbb{Z}/n\mathbb{Z}$ the following implication stands: $$a+b+c+d=0\implies a^7+b^7+c^7+d^7=0.$$

One can note that when $n=7$ we have $(a+b+c+d)^7=a^7+b^7+c^7+d^7$ so the above implication stands.

If $n=2,3$ then $x^7=x$, so it's also true.

For $n=4$ it's false. Counterexample: $(a,b,c,d)=(2,3,3,0)$.

How to find other such $n$ and prove that those are the only ones?

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    $(2,-1,-1,0)$ works as counterexample for a lot of $n$. For instance, it trivially excludes all $n$ greater than $126$ – 2017-01-09
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    @Arthur, yes, true, thank you! This counterexample excludes all cases except $n=2,3,6,7,9,14,18,21,42,63,126$. – 2017-01-09

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The examples $(2,-1,-1,0)$ and $(3,-1,-1,-1)$ show that you need $2^7=2$ and $3^7=3$. This means that $n$ must divide both $2^7-2$ and $3^7-3$, and their greatest common divisor is $42$.

It turns out that $k^7-k$ is divisible by $42$ for every natural $k$ (divisible by $7$ by Fermat's little theorem, by $2$ and $3$ by checking), so the $n$ that work are $1,2,3,6,7,14,21$ and $42$.