For a positive integer $ n,$ $ a_{n}\equal{}n\sqrt{5}\minus{} \lfloor n\sqrt{5}\rfloor$. Compute the maximum value and the minimum value of $ a_{1},a_{2},\ldots ,a_{2009}.$
1. **Understanding the Problem:**
   We need to find the maximum and minimum values of \( a_n = n\sqrt{5} - \lfloor n\sqrt{5} \rfloor \) for \( n = 1, 2, \ldots, 2009 \). This is equivalent to finding the maximum and minimum values of the fractional part of \( n\sqrt{5} \).

2. **Rewriting the Problem:**
   The fractional part of \( n\sqrt{5} \) is given by \( \{ n\sqrt{5} \} \), where \( \{ x \} \) denotes the fractional part of \( x \). We need to analyze the sequence \( \{ n\sqrt{5} \} \) for \( n = 1, 2, \ldots, 2009 \).

3. **Using Properties of the Golden Ratio:**
   Let \( \phi = \frac{1 + \sqrt{5}}{2} \) be the golden ratio. Note that \( \sqrt{5} = 2\phi - 1 \). Therefore, \( n\sqrt{5} = n(2\phi - 1) = 2n\phi - n \).

4. **Modulo 1 Analysis:**
   We are interested in the fractional part of \( 2n\phi - n \). Since \( n \) is an integer, the fractional part of \( 2n\phi - n \) is the same as the fractional part of \( 2n\phi \). Thus, we need to analyze the sequence \( \{ 2n\phi \} \) for \( n = 1, 2, \ldots, 2009 \).

5. **Fibonacci Numbers and Induction:**
   We can use the property of the golden ratio and Fibonacci numbers. By induction, we can show that \( \phi^{k+1} = F_k\phi + F_{k-1} \), where \( F_k \) are the Fibonacci numbers with \( F_0 = 0 \) and \( F_1 = 1 \). This implies that \( F_k\phi \equiv \phi^{k+1} \pmod{1} \).

6. **Behavior of \( \phi^k \):**
   The term \( \left( \frac{1 + \sqrt{5}}{2} \right)^k + \left( \frac{1 - \sqrt{5}}{2} \right)^k \) is an integer for all positive integers \( k \), and \( \left( \frac{1 - \sqrt{5}}{2} \right)^k \) tends to 0 as \( k \to \infty \). Therefore, \( \phi^k \) gets closer to an integer as \( k \) increases.

7. **Maximizing and Minimizing the Fractional Part:**
   To maximize \( \{ 2n\phi \} \), we want \( 2n\phi \) to be as close to 1 as possible. To minimize \( \{ 2n\phi \} \), we want \( 2n\phi \) to be as close to 0 as possible.

8. **Using Fibonacci Representation:**
   Every positive integer can be written as the sum of distinct, nonconsecutive Fibonacci numbers. Let \( 2n = F_{k_1} + F_{k_2} + \cdots + F_{k_r} \) with \( k_1 < k_2 < \cdots < k_r \) and no two consecutive. Then \( 2n\phi = \phi^{k_1+1} + \phi^{k_2+1} + \cdots + \phi^{k_r+1} \pmod{1} \).

9. **Maximizing \( \{ 2n\phi \} \):**
   To maximize \( \{ 2n\phi \} \), we choose the largest odd \( k_1 \) possible, then even \( k_2, k_3, \ldots \) to push the number closer to 1. The best we can do is \( 2n = F_{17} = 2584 \), so \( n = 1292 \).

10. **Minimizing \( \{ 2n\phi \} \):**
    To minimize \( \{ 2n\phi \} \), we choose the largest even \( k_1 \) possible, then odd \( k_2, k_3, \ldots \) to push the number closer to 0. We can't use \( 2n = F_{16} = 1597 \) since 1597 is odd, so we use \( 2n = F_{14} + F_{17} = 610 + 2584 = 3194 \), so \( n = 1597 \).

The final answer is \(\boxed{1292}\) for the maximum and \(\boxed{1597}\) for the minimum.