Let $x_1=a$, $x_2=b-a$, $x_3=c-b$, $x_4=d-c$, and $x_5=11-d$. Then, it holds that $$x_1+x_2+x_3+x_4+x_5=11$$

There are $\binom{11-1}{5-1}=210$ positive integer solutions to this equation. However, following possibilities need to be excluded where $$a+d > b+c\implies d-c > b-a \implies x_4 > x_2$$

By symmetry, there are as many quadruples where $x_4 > x_2$ as those where $x_2 > x_4$. Meanwhile, those cases where $x_2=x_4$ can be manually counted:

•   $x_2=x_4=1\implies x_1+x_3+x_5=9\implies \binom{9-1}{3-1}=28$
•   $x_2=x_4=2\implies x_1 + x_3+x_5=7\implies \binom{7-1}{3-1}=15$
•   $x_2=x_4=3\implies x_1+x_3+x_5=5\implies \binom{5-1}{3-1}=6$
•   $x_2=x_4=4\implies x_1+x_3+x_5=3\implies 1$

Therefore, the number of interesting quadruples is $$(210-28-15-6-1)\div 2 =\boxed{80} $$