B. If the positive integer $n$ satisfies that the sum of the digits of $n$ is 2014, and the sum of the digits of $5n$ is 1007, then ( ).
(A) $n$ must be odd
(B) $n$ must be even
(C) $n$ can be odd or even
(D) $n$ does not exist
B. B.

For each positive integer $m$, let the sum of its digits be denoted as $S(m)$.
If a positive integer $n$ satisfies the condition, then
$$
S(10 n)=S(n)=S(5 n)+S(5 n) \text {. }
$$

This indicates that no carry occurs when $5 n$ and $5 n$ are added.
Since the last digit of $5 n$ is 0 or 5, and when the last digit of $5 n$ is 5, a carry occurs in the units place of $5 n+5 n$, thus, the last digit of $5 n$ must be 0. Therefore, $n$ can only be an even number.

Moreover, such an $n$ does indeed exist, for example, $n=\frac{22 \cdots 2}{1007 \uparrow}$, in this case, $S(n)=2014$, and $5 n=\frac{11 \cdots 10}{1007 \uparrow}$, such that
$$
S(5 n)=1007 .
$$