6. There are four racing boats numbered (1), (2), (3), and (4), with speeds of $v_{1}$, $v_{2}$, $v_{3}$, and $v_{4}$ kilometers per hour, respectively, and they satisfy $v_{1}>v_{2}>v_{3}>v_{4}>v_{\text {water }}>0$, where $v_{\text {water }}$ is the speed of the river current (kilometers/hour). They are participating in a pursuit race with the following rules:
(1) The four boats start from the same starting line simultaneously. Boats (1), (2), and (3) head upstream, while boat (4) heads downstream.
(2) After 1 hour, boats (1), (2), and (3) turn around and chase boat (4). The first one to catch up with boat (4) is the champion.
According to these rules, the champion is ( ).
(A) Boat (1)
(B) Boat (2)
(C) Boat (3)
(D) Boats (1), (2), and (3) tie
6. A.

1 hour after departure, the distances between boats (1), (2), (3) and boat (4) are
$$
S_{i}=\left[\left(v_{i}-v_{\text {t }}\right)-\left(v_{k}+v_{i}\right)\right] \times 1=v_{i}+v_{i} .
$$

The time it takes for each boat to catch up with boat (4) is
$$
\begin{array}{l}
t_{i}=\frac{v_{i}+v_{4}}{\left(v_{i}+v_{\text {water }}\right)-\left(v_{\text {water }}+v_{4}\right)} \\
=\frac{v_{i}+v_{4}}{v_{i}-v_{4}}=1+\frac{2 v_{4}}{v_{i}-v_{4}} . \\
\end{array}
$$

For $v_{1}>v_{2}>v_{3}>v_{4}$, we have
$$
t_{1}<t_{2}<t_{3},
$$

which means boat (1) takes the least time to catch up with boat (4).
Explanation: If boat (4) is replaced by a buoy, $v_{4}=0$, then $t_{i}=1$, meaning boats (1), (2), and (3) arrive simultaneously.