The captain and his three sailors get $2009$ golden coins with the same value . The four people decided to divide these coins by the following rules :
sailor $1$,sailor $2$,sailor $3$ everyone write down an integer $b_1,b_2,b_3$ , satisfies $b_1\ge b_2\ge b_3$ , and ${b_1+b_2+b_3=2009}$; the captain dosen't know what the numbers the sailors have written . He divides $2009$ coins into $3$ piles , with number of coins: $a_1,a_2,a_3$ , and $a_1\ge a_2\ge a_3$ . For sailor $k$ ($k=1,2,3$) , if $b_k<a_k$ , then he can take $b_k$ coins from the $k$th pile ; if $b_k\ge a_k$ , then he can't take any coins away . At last , the captain own the rest of the coins .If no matter what the numbers the sailors write , the captain can make sure that he always gets $n$ coins . 
Find the largest possible value of $n$ and prove your conclusion .
To solve this problem, we need to ensure that the captain always gets a certain number of coins, regardless of the integers \( b_1, b_2, b_3 \) chosen by the sailors. Let's analyze the problem step-by-step.

1. **Understanding the Constraints:**
   - The sailors write down integers \( b_1, b_2, b_3 \) such that \( b_1 \ge b_2 \ge b_3 \) and \( b_1 + b_2 + b_3 = 2009 \).
   - The captain divides the 2009 coins into three piles \( a_1, a_2, a_3 \) such that \( a_1 \ge a_2 \ge a_3 \).
   - For each sailor \( k \) (where \( k = 1, 2, 3 \)), if \( b_k < a_k \), the sailor takes \( b_k \) coins from the \( k \)-th pile. If \( b_k \ge a_k \), the sailor takes no coins from the \( k \)-th pile.

2. **Analyzing the Worst-Case Scenario:**
   - The worst-case scenario for the captain is when the sailors choose \( b_1, b_2, b_3 \) in such a way that maximizes the number of coins they can take.
   - Consider the case where \( b_1 = 2009 \), \( b_2 = 0 \), and \( b_3 = 0 \). In this case, \( b_1 \ge a_1 \), so the first sailor cannot take any coins. The same applies to the other sailors since \( b_2 \ge a_2 \) and \( b_3 \ge a_3 \).

3. **Ensuring the Captain Gets Coins:**
   - To ensure the captain always gets a certain number of coins, we need to find the maximum value of \( n \) such that the captain can always secure \( n \) coins regardless of the sailors' choices.
   - Let's consider the distribution of coins in the piles. If the captain distributes the coins as \( a_1 = a_2 = a_3 = 669 \) (since \( 2009 \div 3 \approx 669 \)), then:
     - If \( b_1, b_2, b_3 \) are such that \( b_1 \ge 669 \), \( b_2 \ge 669 \), and \( b_3 \ge 669 \), then no sailor can take any coins.
     - If \( b_1, b_2, b_3 \) are such that \( b_1 < 669 \), \( b_2 < 669 \), and \( b_3 < 669 \), then the sailors can take at most \( b_1 + b_2 + b_3 = 2009 \) coins, but this is not possible since \( b_1 + b_2 + b_3 = 2009 \).

4. **Finding the Largest Possible Value of \( n \):**
   - To maximize \( n \), we need to consider the worst-case scenario where the sailors take the maximum number of coins possible.
   - If the captain distributes the coins as \( a_1 = 1005 \), \( a_2 = 1004 \), and \( a_3 = 0 \), then:
     - If \( b_1 = 2009 \), \( b_2 = 0 \), and \( b_3 = 0 \), the first sailor cannot take any coins.
     - If \( b_1 = 1005 \), \( b_2 = 1004 \), and \( b_3 = 0 \), the first two sailors take all the coins, leaving the captain with 0 coins.

5. **Conclusion:**
   - The captain can ensure he always gets at least \( n = 669 \) coins by distributing the coins as \( a_1 = 669 \), \( a_2 = 669 \), and \( a_3 = 671 \).
   - In this distribution, the sailors can take at most \( 669 + 669 + 671 = 2009 \) coins, but since \( b_1 + b_2 + b_3 = 2009 \), the captain will always get at least 669 coins.

The final answer is \( \boxed{669} \).