For all positive reals $ a$, $ b$, and $ c$, what is the value of positive constant $ k$ satisfies the following inequality?
$ \frac{a}{c\plus{}kb}\plus{}\frac{b}{a\plus{}kc}\plus{}\frac{c}{b\plus{}ka}\geq\frac{1}{2007}$ .
1. We start with the given inequality:
   \[
   \frac{a}{c + kb} + \frac{b}{a + kc} + \frac{c}{b + ka} \geq \frac{1}{2007}
   \]
   We need to find the value of the positive constant \( k \) that satisfies this inequality for all positive reals \( a \), \( b \), and \( c \).

2. Let's test the inequality with \( a = b = c = 1 \):
   \[
   \frac{1}{1 + k} + \frac{1}{1 + k} + \frac{1}{1 + k} = \frac{3}{1 + k}
   \]
   We need this to be at least \(\frac{1}{2007}\):
   \[
   \frac{3}{1 + k} \geq \frac{1}{2007}
   \]
   Solving for \( k \):
   \[
   3 \geq \frac{1 + k}{2007}
   \]
   \[
   3 \times 2007 \geq 1 + k
   \]
   \[
   6021 \geq 1 + k
   \]
   \[
   k \leq 6020
   \]

3. Now, we need to verify that \( k = 6020 \) satisfies the inequality for all positive reals \( a \), \( b \), and \( c \). We use the Titu's Lemma (or Cauchy-Schwarz inequality in Engel form):
   \[
   \sum_{cyc} \frac{a}{c + kb} \geq \frac{(a + b + c)^2}{a(c + kb) + b(a + kc) + c(b + ka)}
   \]
   Simplifying the denominator:
   \[
   a(c + kb) + b(a + kc) + c(b + ka) = ac + akb + ba + bkc + cb + cka = (k + 1)(ab + bc + ca)
   \]
   Therefore:
   \[
   \sum_{cyc} \frac{a}{c + kb} \geq \frac{(a + b + c)^2}{(k + 1)(ab + bc + ca)}
   \]
   We need:
   \[
   \frac{(a + b + c)^2}{(k + 1)(ab + bc + ca)} \geq \frac{1}{2007}
   \]
   Substituting \( k = 6020 \):
   \[
   \frac{(a + b + c)^2}{6021(ab + bc + ca)} \geq \frac{1}{2007}
   \]
   Simplifying:
   \[
   2007(a + b + c)^2 \geq 6021(ab + bc + ca)
   \]
   Dividing both sides by 2007:
   \[
   (a + b + c)^2 \geq 3(ab + bc + ca)
   \]
   This is true by the AM-GM inequality, which states:
   \[
   (a + b + c)^2 \geq 3(ab + bc + ca)
   \]

Since all steps are verified, the value of \( k \) that satisfies the inequality is \( k = 6020 \).

The final answer is \( \boxed{6020} \).