Let the G.P. have first term $a_1=a$ and common ratio $r$.

Then $a_2=ar,\, a_3=ar^2,\, a_4=ar^3.$

Use the given conditions:

$a_1+a_2=a+ar=a(1+r)=4 \quad ...(1)$

$a_3+a_4=ar^2+ar^3=ar^2(1+r)=16 \quad ...(2)$

Divide (2) by (1):

$\frac{ar^2(1+r)}{a(1+r)}=\frac{16}{4}$ $r^2=4$ $r=\pm 2$

Now use the condition $a_1<0$.

From (1), $a=\frac{4}{1+r}.$

• If $r=2$, then $a=\frac{4}{3}>0,$ which contradicts $a_1<0$.

• If $r=-2$, then $a=\frac{4}{1-2}=-4,$ which satisfies $a_1<0$.

Hence, $a_1=-4, \quad r=-2.$

Sum of first $9$ terms of a G.P. is

$S_9=a\frac{1-r^9}{1-r}.$

Substitute $a=-4$, $r=-2$:

$S_9=-4\cdot \frac{1-(-2)^9}{1-(-2)}$ $=-4\cdot \frac{1-(-512)}{3}$ $=-4\cdot \frac{513}{3}$ $=-4\cdot 171$ $=-684.$

Given $\sum_{i=1}^9 a_i = 4\lambda,$ so $4\lambda=-684$ $\lambda=-171.$

Therefore, the correct option is: $\boxed{-171}$ which is option B.