Find all $n\in\mathbb{N}$ such that there exist $a, b, c, d\in\mathbb{C}$ satisfying $a^n=b^n=c^n=d^n=1$ and $a+b+c+d=1$.
What I already know:
For $n=1,2,4$ is false. It's easy to check.
For $n=3k,k\in\mathbb{N}$ is true by choosing $d=1$ and $a,b,c$ are three different root of $z^3=1$
How about $n=3k+1$ and $n=3k+2$ for $n\geq5$?
Denoting $\zeta_n:=\exp(2\pi i/n)$, the problem can be reformulated as follows:
Find all $n\in\mathbb{N}$ such that there exist $a_1,a_2,a_3,a_4\in\mathbb{Z}$ satisfying $\zeta_n^{a_1}+\zeta_n^{a_2}+\zeta_n^{a_3}+\zeta_n^{a_4}=1.$
The answer is given by $\color{blue}{n\in(3\mathbb{N})\cup(10\mathbb{N})}$; we're going to prove it. Solutions $(a_1,a_2,a_3,a_4)$ of this problem for $n=3$ (as you noted) and $n=10$ are given by $(0,0,1,2)$ and $(1,3,7,9)$, respectively.
Now suppose $(a_1,a_2,a_3,a_4)$ is a solution for $n\in\mathbb{N}$ with $3\not\mid n$ and $10\not\mid n$.
So, if $\Phi(\zeta):=1-\sum_{k=1}^{4}\zeta^{a_k}$, we have $\Phi(\zeta_n)=0$. Then $\Phi(\zeta_n^r)=0$ for each integer $r$ coprime to $n$ (we may assume $0\leqslant a_k\leqslant n-1$, making $\Phi(\zeta)$ a polynomial over $\mathbb{Q}$; then $\zeta_n$ is a common root of $\Phi(\zeta)$ and the cyclotomic polynomial $\Phi_n(\zeta)$, which is irreducible over $\mathbb{Q}$, hence is a factor of $\Phi(\zeta)$, and has roots at $\zeta=\zeta_n^r$).
Summing over $1\leqslant r\leqslant n$ (still under $\gcd(n,r)=1$) and using this formula, we obtain $$\color{blue}{\sum_{k=1}^{4}\rho(d_k)=1},\quad\rho(d):=\frac{\mu(d)}{\varphi(d)},\quad d_k:=\frac{n}{\gcd(n,a_k)}.$$ If, say, $d_4=1$, then $n\mid a_4$ and $\zeta_n^{a_1}+\zeta_n^{a_2}+\zeta_n^{a_3}=0$, which gives $$1+\zeta_n^{a_2-a_1}=-\zeta_n^{a_3-a_1}\implies(1+\zeta_n^{a_2-a_1})^n=(-1)^n;$$ denoting $s:=a_2-a_1$, we get $(\zeta_{2n}^s+\zeta_{2n}^{-s})^n=(-1)^{n+s}$, i.e. $\cos(s\pi/n)=\pm 1/2$, impossible since $3\not\mid n$. Thus, $d_k\neq 1$ for each $k$. Also, $3\not\mid d_k$ and $10\not\mid d_k$, which implies $\color{blue}{\rho(d_k)\leqslant 1/6}$. Contradiction.