3. Find all positive integers $n \geqslant 4$ such that for any distinct and non-zero complex numbers $a, b, c$, if
$$
(a-b)^{n}+(b-c)^{n}+(c-a)^{n}=0,
$$

then $a, b, c$ are the complex coordinates of the vertices of an equilateral triangle.
3. Let's assume $a-b=1, b-c=x, c-a=-1-x$.

Let the three roots of $z^{3}-1=0$ be $1, \omega, \omega^{2}$, where $\omega=-\frac{1}{2}+\frac{\sqrt{3}}{2}i$.

If $a, b, c$ are the complex coordinates of the vertices of an equilateral triangle, then $x=\omega$ or $x=\omega^{2}$.
From the problem, we have
$$
\begin{array}{l}
0=(a-b)^{n}+(b-c)^{n}+(c-a)^{n} \\
=1+x^{n}+(-1)^{n}(x+1)^{n} .
\end{array}
$$

If equation (1) has a solution not equal to 1, and not equal to $\omega$ or $\omega^{2}$, then there exist $a, b, c$ that are not the complex coordinates of the vertices of an equilateral triangle.
When $2 \nmid n (n \geqslant 5)$, let's assume $n=2 k+1$, then
$$
\begin{array}{l}
1+x^{n}+(-1)^{n}(x+1)^{n} \\
=1+x^{2 k+1}-(x+1)^{2 k+1} \\
=-x\left((2 k+1) x^{2 k-1}+\cdots+(2 k+1)\right) .
\end{array}
$$

Clearly, the equation
$$
(2 k+1) x^{2 k-1}+\cdots+(2 k+1)=0
$$

has a non-zero real root.
Thus, there exist $a, b, c$ that are not the complex coordinates of the vertices of an equilateral triangle.
When $2 \mid n (n \geqslant 4)$, let's assume $n=2 k$, then
$$
\begin{array}{l}
1+x^{n}+(-1)^{n}(x+1)^{n} \\
=1+x^{2 k}+(x+1)^{2 k} .
\end{array}
$$

Clearly, the equation
$$
1+x^{2 k}+(x+1)^{2 k}=0
$$

has no zero root.
Thus, if $a, b, c$ are the complex coordinates of the vertices of an equilateral triangle, then
$$
1+x^{2 k}+(x+1)^{2 k}=0
$$

has roots $\omega, \omega^{2}$.
Since $\omega, \omega^{2}$ are conjugates, let's assume
$$
\begin{array}{l}
1+x^{2 k}+(x+1)^{2 k} \\
=2\left((x-\omega)\left(x-\omega^{2}\right)\right)^{k} \\
=2\left(x^{2}+x+1\right)^{k} .
\end{array}
$$

Considering the coefficient of $x^{2}$ on both sides of equation (2), the coefficient of $x^{2}$ on the left side is $\mathrm{C}_{2 k}^{2}=k(2 k-1)$, and the coefficient of $x^{2}$ on the right side is
$$
\begin{array}{l}
2\left(\mathrm{C}_{k}^{2}+\mathrm{C}_{k}^{1}\right)=k(k+1) . \\
\text { By } k(2 k-1)=k(k+1) \\
\Rightarrow k=2(k=0 \text { is discarded }) . \\
\text { When } k=2, n=4, \text { at this time, } \\
1+x^{n}+(-1)^{n}(x+1)^{n} \\
=1+x^{4}+(x+1)^{4} \\
=2\left(x^{2}+x+1\right)^{2},
\end{array}
$$

When $k=2$, $n=4$, at this time,

it satisfies the problem.
In summary, $n=4$ is the only positive integer that satisfies the condition.