Count 'em All 10!

From this note , we know that the number of quadrilaterals in an $a\times b$ grid is ${a+1\choose2}\times{b+1\choose2}=\frac{ab(a+1)(b+1)}{4}$

Now, by substituting $a=20$ and $b=30$ , we would get $97650$ .

Thè number of quadrilaterals in $(20×30)$ grid

= $\displaystyle{\sum_{n=1}^{20}} × \displaystyle{\sum_{n=1}^{30}}$

= $210 × 465$
= $\boxed {97650}$ $_\square$