These sorts of geometry problems are pretty popular...

First, a confession. The main focus of this problem isn't geometry. It's actually algebra! Let me explain why:

Let's let $a_1, a_2, \dots, a_6$ represent the lengths of the smaller segments of $AD$ going from top to bottom, and let $b_1, b_2, \dots, b_6$ represent the lengths of the smaller segments of $AB$ going from right to left. From the problem statement, we know that $a_{i + 1} \geq a_i$ and $b_{i + 1} \geq b_i$ for $i = 1, 2, 3, 4, 5.$

By Chebyshev's Sum Inequality ,

$\dfrac{1}{6} \left ( \sum_{i = 1}^6 a_i \right ) \left ( \sum_{i = 1}^6 b_i \right ) \geq \sum_{i = 1}^6 a_i b_{7 - i}.$

However, notice that the right hand side of this inequality is actually the area of the blue region, which is $a_1b_6 + a_2b_5 + \dots + a_6b_1.$ Furthermore, since $\displaystyle \sum_{i = 1}^6 a_i = AD$ and $\displaystyle \sum_{i = 1}^6 b_i = AB,$ the left hand side of the inequality is equal to $\frac{1}{6}$ of the area of $ABCD.$

This leads us to conclude that the blue region has at most as much area as one-sixth of $ABCD.$ Since one-fifth of $ABCD$ has strictly greater area than one-sixth of $ABCD,$ we conclude that $\boxed{\frac{1}{5} \text{ of } ABCD}$ always has a greater area than the blue region, no matter how we partition $ABCD$ according to the given rules.