4. Given that $n$ is even. If the sum of the first $n$ positive integers starting from 1 has a units digit of 8, then the units digit of the sum of the next $n$ positive integers is ( ).
(A) 6
(B) 4
(C) 2
(D) 0
4. C.

Let $n=2k$, and let $S$ be the sum of the first $n$ positive integers, and $D$ be the sum of the next $n$ positive integers (from $2k+1$ to $4k$).
$$
\begin{array}{l}
\text { Then } S=k(2k+1), \\
D=\frac{2k(2k+1+4k)}{2}=(6k+1)k .
\end{array}
$$

The last digits of $S$ and $D$ depend only on the last digit of $k$. According to the problem, $k$ should end in 6.
Thus, the last digit of $D$ is the last digit of $(6 \times 6+1) \times 6$, which is 2.