Four. (25 points) A captain and three sailors together received 2,009 coins of the same denomination. The four agreed to distribute the coins according to the following rules: Sailor 1, Sailor 2, and Sailor 3 each write down a positive integer, denoted as \( b_{1}, b_{2}, b_{3} \), satisfying \( b_{1} \geqslant b_{2} \geqslant b_{3} \), and \( b_{1} + b_{2} + b_{3} = 2009 \); the captain, without knowing the numbers written by the sailors, divides the 2009 coins into 3 piles, with the quantities of each pile being \( a_{1}, a_{2}, a_{3} \), and \( a_{1} \geqslant a_{2} \geqslant a_{3} \). For sailor \( k \) (where \( k = 1, 2, 3 \)), if \( b_{k} < a_{k} \), they can take \( b_{k} \) coins from the \( k \)-th pile; otherwise, they cannot take any. All remaining coins go to the captain. If the captain can always ensure that he gets \( n \) coins regardless of how the three sailors write their numbers, determine the maximum value of \( n \) and prove your conclusion.
（Zhang Limin, Contributed）
Four, the maximum value is 673.
First, the captain can ensure getting no less than 673 gold coins.

In fact, when the captain divides the gold coins into 3 piles with numbers $671, 670, 668$,
(1) If $b_{1} \geqslant 671$, then the captain can get no less than $671+2=673$ coins;
(2) If $b_{1} < 671$, then the captain can get no less than $670+3=673$ coins.

Second, the captain cannot ensure getting more than 673 gold coins.
In fact,
(1) If $a_{1} \leqslant 671$, then $a_{2} \leqslant 671, a_{3} \geqslant 667$. When $b_{1}=a_{1}+2, b_{2}=a_{2}-1, b_{3}=a_{3}-1$, the captain can get at most $671+2=673$ coins;
(2) If $a_{1}>671$, then because
$$
a_{3} \leqslant \frac{2009-a_{1}}{2} \leqslant \frac{1337}{2}=668.5 \text {, }
$$

thus $a_{3} \leqslant 668$.
(i) When $a_{2}-a_{3} \geqslant 3$, if $b_{1}=a_{1}-1, b_{2}=$ $a_{2}-1, b_{3}=a_{3}+2$, then the captain can get at most $a_{3}+2 \leqslant$ $668+2=670$ coins;
(ii) When $a_{2}-a_{3} \leqslant 2$, if $a_{3}=1$, then
$$
a_{2} \leqslant 3, a_{1}-a_{2} \geqslant 2002 \text {, }
$$

when $b_{1}=a_{1}-2, b_{2}=a_{2}+1, b_{3}=a_{3}+1$, the captain can get at most $2+3+1=6$ coins;
$$
\begin{array}{l}
\text { If } a_{3}>1 \text {, then } \\
2 a_{2}=\left(a_{2}+a_{3}\right)+\left(a_{2}-a_{3}\right) \\
\leqslant 2009-672+2=1339 \\
\Rightarrow a_{2} \leqslant 669 .
\end{array}
$$

If $a_{3}>1$, then

Therefore, $a_{1}-a_{2} \geqslant 672-669=3$, when $b_{1}=a_{1}$ $-1, b_{2}=a_{2}+2, b_{3}=a_{3}-1$, the captain can get at most $a_{2}+2 \leqslant 671$ coins.