4. Let $a$, $b$, $c$ be positive integers, and satisfy
$$
a^{2}+b^{2}+c^{2}-a b-b c-c a=19 \text {. }
$$

Then the minimum value of $a+b+c$ is $\qquad$
4. 10 .

By the cyclic symmetry of $a, b, c$, without loss of generality, assume $a \geqslant b \geqslant c$.
Let $a-b=m, b-c=n$, then $a-c=m+n$, and
$$
\begin{array}{l}
a^{2}+b^{2}+c^{2}-a b-b c-c a \\
=\frac{1}{2}\left[m^{2}+n^{2}+(m+n)^{2}\right],
\end{array}
$$

which is $m^{2}+m n+n^{2}-19=0$.
For the equation in $m$,
$$
\Delta=n^{2}-4\left(n^{2}-19\right) \geqslant 0 \text {. }
$$

Thus, $0 \leqslant n \leqslant \frac{2 \sqrt{57}}{3}0)$ has no solution.

Therefore, when $a=6, b=3, c=1$, $a+b+c$ achieves its minimum value, which is 10.