Proof: First, if two judges give the same evaluation to a contestant, we call it an "agreement." From the given information, any two judges can have at most $k$ "agreements." Thus,
the total number of "agreements" $\leqslant k \mathrm{C}_{b}^{2}$.
On the other hand, for any one contestant, suppose $A$ judges pass them, and $B$ judges fail them, where $A+B=b$. Then, for this contestant, the number of "agreements" related to them is
$$\begin{array}{l} \mathrm{C}_{A}^{2}+\mathrm{C}_{B}^{2}=\frac{A^{2}-A+B^{2}-B}{2} \\ =\frac{(A+B)^{2}+(A-B)^{2}-2(A+B)}{4} \\ =\frac{b^{2}-2 b+(A-B)^{2}}{4} . \end{array}$$

Since $b$ is odd, $A-B$ is also odd. Therefore, $|A-B| \geqslant 1$. Thus,
$$\mathrm{C}_{A}^{2}+\mathrm{C}_{B} \geqslant \frac{\dot{b}^{2}-2 b+1}{4}=\left(\frac{b-1}{2}\right)^{2} .$$

Therefore,
the total number of "agreements" $\geqslant a\left(\frac{b-1}{2}\right)^{2}$.
From (1) and (2), we get $a\left(\frac{b-1}{2}\right)^{2} \leqslant k \mathrm{C}_{b}^{2}$,

i.e., $\frac{k}{a} \geqslant \frac{b-1}{2 b}$.