Let $a$, $b$, $c$, and $d$ be positive real numbers such that 
\[\begin{array}{c@{\hspace{3pt}} c@{\hspace{3pt}} c@{\hspace{3pt}} c@{\hspace{3pt}}c}a^2+b^2&=&c^2+d^2&=&2008,\\ ac&=&bd&=&1000.\end{array}\]If $S=a+b+c+d$, compute the value of $\lfloor S\rfloor$.
1. Given the equations:
   \[
   a^2 + b^2 = c^2 + d^2 = 2008 \quad \text{and} \quad ac = bd = 1000
   \]
   we need to find the value of \( \lfloor S \rfloor \) where \( S = a + b + c + d \).

2. First, consider the product of the sums of squares:
   \[
   (a^2 + b^2)(c^2 + d^2) = 2008^2
   \]
   Expanding this product, we get:
   \[
   a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2 = 2008^2
   \]

3. Using the given \( ac = bd = 1000 \), we can rewrite the terms:
   \[
   (ac)^2 + (ad)^2 + (bc)^2 + (bd)^2 = 1000^2 + (ad)^2 + (bc)^2 + 1000^2
   \]
   Simplifying, we get:
   \[
   1000000 + (ad)^2 + (bc)^2 + 1000000 = 2008^2
   \]
   \[
   2000000 + (ad)^2 + (bc)^2 = 2008^2
   \]

4. Since \( ac = bd = 1000 \), we can use the identity for the sum of squares:
   \[
   (ad)^2 + (bc)^2 = (ad + bc)^2 - 2(ad \cdot bc)
   \]
   Given \( ad + bc = 2008 \), we substitute:
   \[
   (ad + bc)^2 = 2008^2
   \]
   \[
   (ad + bc)^2 = 2008^2
   \]

5. This implies:
   \[
   ad + bc = 2008
   \]

6. Now, we use the fact that \( a^2 + b^2 = c^2 + d^2 = 2008 \) and \( ad + bc = 2008 \):
   \[
   a^2 + b^2 + c^2 + d^2 - 2(ad + bc) = 0
   \]
   \[
   (a - d)^2 + (b - c)^2 = 0
   \]
   This implies:
   \[
   a = d \quad \text{and} \quad b = c
   \]

7. Given \( a = d \) and \( b = c \), we substitute back into the original equations:
   \[
   a^2 + b^2 = 2008 \quad \text{and} \quad ab = 1000
   \]

8. Solving for \( a \) and \( b \):
   \[
   a^2 + b^2 = 2008
   \]
   \[
   ab = 1000
   \]

9. Let \( x = a + b \) and \( y = ab \):
   \[
   x^2 - 2y = 2008
   \]
   \[
   y = 1000
   \]
   \[
   x^2 - 2000 = 2008
   \]
   \[
   x^2 = 4008
   \]
   \[
   x = \sqrt{4008}
   \]

10. Since \( a + b = c + d = \sqrt{4008} \), we have:
    \[
    S = a + b + c + d = 2\sqrt{4008}
    \]

11. Finally, we compute \( \lfloor S \rfloor \):
    \[
    \lfloor 2\sqrt{4008} \rfloor
    \]
    Approximating \( \sqrt{4008} \approx 63.34 \):
    \[
    2 \times 63.34 = 126.68
    \]
    \[
    \lfloor 126.68 \rfloor = 126
    \]

The final answer is \( \boxed{126} \).