4. If the inequality $2|x-1|+3|x-3| \leqslant a$ has a solution, then the minimum value of the real number $a$ is ( ).
(A) 1
(B) 2
(C) 4
(D) 6
4. C.

By the geometric meaning of absolute value,
$$
\begin{array}{l}
2|x-1|+3|x-3| \\
=2(|x-1|+|x-3|)+|x-3| \\
\geqslant 2 \times 2+0=4,
\end{array}
$$

When $x=3$, the equality holds.