1708
\hspace{-16}$Here Integral point must satisfy the Condition \\\\ $\mathbf{x,y<0}$ and $\mathbf{x+y<21}$\\\\ So $\mathbf{0<x,y<21}$ and $\mathbf{x\in\mathbb{Z^{+}}}$\\\\ So $\mathbf{x,y\in \left\{1,2,3,4,.............,20\right\}}$\\\\ So If $\mathbf{x=1}$. Then $\mathbf{y\in \left\{1,2,3,4,........,19\right\}}$(bcz $\mathbf{x+y<21}$)\\\\ So Total $\mathbf{19}$ Integers pairs.\\\\ Similarly\\\\ So If $\mathbf{x=2}$. Then $\mathbf{y\in \left\{1,2,3,4,........,18\right\}}$(bcz $\mathbf{x+y<21}$)\\\\ So Total $\mathbf{18}$ Integers pairs.\\\\ So If $\mathbf{x=3}$. Then $\mathbf{y\in \left\{1,2,3,4,........,17\right\}}$(bcz $\mathbf{x+y<21}$)\\\\ So Total $\mathbf{18}$ Integers pairs.\\\\ ..................\\\\ ...................\\\\ ....................\\\\ So If $\mathbf{x=19}$. Then $\mathbf{y\in \left\{1\right\}}$(bcz $\mathbf{x+y<21}$)\\\\
\hspace{-18}$So Total $\mathbf{1}$ Integers pairs.\\\\ So If $\mathbf{x=20}$. Then $\mathbf{y\in \left\{0\right\}}$(bcz $\mathbf{x+y<21}$)\\\\ So Total $\mathbf{0}$ Integers pairs.\\\\ So Toal Integers points within The $\mathbf{\triangle OAB}$ is\\\\ $\mathbf{=19+18+17+..........+1+0=\frac{19.20}{2}=190}$
