Let equation of line is
$\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1\Rightarrow \frac{2}{\mathrm{a}}+\frac{2}{\mathrm{b}}=1$
$\Rightarrow \hspace{0.17em}2\left(\mathrm{b}+\mathrm{a}\right)=\mathrm{a}\mathrm{b}$ ... $\left(1\right)$
Case-1: If both intercepts are positive
Then, $\frac{1}{2}\mathrm{a}\mathrm{b}=4\Rightarrow \mathrm{a}\mathrm{b}=8\Rightarrow b+a=4$
$\Rightarrow \hspace{0.17em}\mathrm{a}\left(4-\mathrm{a}\right)8\Rightarrow {\mathrm{a}}^2-4\mathrm{a}+8=0$ $\Rightarrow a\mathrm{}\left(4-a\right)=8$
$\Rightarrow$ No value of $\mathrm{a}$ is possible (Discriminant is negative)
Case-2: If intercepts are of opposite sign, then $-\frac{1}{2}\mathrm{a}\mathrm{b}=4\Rightarrow \mathrm{a}\mathrm{b}=-8$
$\Rightarrow \mathrm{b}+\mathrm{a}=\frac{-8}{2}=-4$
$\Rightarrow \mathrm{a}\left(-4-\mathrm{a}\right)=-8\Rightarrow {\mathrm{a}}^2+4\mathrm{a}-8=0$
$\Rightarrow$ $\mathrm{a}$ is possible hence $\mathrm{b}$ is possible
$\Rightarrow$ the sum of intercepts is equal to $-4$