4. Let three different lines be
$$
\begin{array}{l}
l_{1}: a x+2 b y+3(a+b+1)=0, \\
l_{2}: b x+2(a+b+1) y+3 a=0, \\
l_{3}:(a+b+1) x+2 a y+3 b=0 .
\end{array}
$$

Then the necessary and sufficient condition for them to intersect at one point is
4. $a+b=-\frac{1}{2}$.

Let $c=a+b+1$, and suppose three lines intersect at point $(x, y)$. Then
$$
\left\{\begin{array}{l}
a x+2 b y+3 c=0, \\
b x+2 c y+3 a=0, \\
c x+2 a y+3 b=0 .
\end{array}\right.
$$

Eliminating $x$ and $y$ yields
$$
\begin{array}{l} 
a^{3}+b^{3}+c^{3}-3 a b c=0 \\
\quad \Rightarrow(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-a c\right)=0 \\
\Rightarrow(2 a+2 b+1)\left((a-b)^{2}+(b+1)^{2}+(a+1)^{2}\right) \\
\quad=0 .
\end{array}
$$

When $(a-b)^{2}+(b+1)^{2}+(a+1)^{2}=0$, solving gives $a=b=-1$, which is not valid and should be discarded;
When $2 a+2 b+1=0$, solving gives $a+b=-\frac{1}{2}$.
Conversely, when $a+b=-\frac{1}{2}$, the system of equations has a solution.