Assume the contrary, i.e. there aren't lines $l$ and $l'$, satisfying the conditions 1) and 2).
Let's choose an arbitrary $l \in L$. There are two points $a, b \in l$ such that $a, b$ are intersection points of $l$ with lines from $L$, and all intersection points of the line $l$ with lines from $L$ lie on the segment $[a, b]$. This $[a, b]$ is called a big segment (of the line $l$).
There is another line, say $l'$, that passes through the endpoint $a$ of the big segment $[a, b]$. Then, by the assumption, this $a$ will be an endpoint of the big segment of the line $l'$.
Denote by $A$ the set of all endpoints of big segments; and if $[a, b]$ is a big segment then we write $aRb$ and say that $a$ and $b$ are in the relation $\mathcal{R}$.
Lemma 1. Let $\mathcal{R}$ be an irreflexive, symmetric relation on the finite set $A$, i.e. $x\mathcal{R}y \Rightarrow x \neq y \& y\mathcal{R}x$. Suppose that for any point $a \in A$ there are exactly two points $b \in A$ such that $a\mathcal{R}b$. Then $A$ is partitioned into a finite $\mathcal{R}$-cycles of the length $\ge 3$.
Proof. is well-known.
But in our situation there is an unique $\mathcal{R}$-cycle of the odd length.
Lemma 2. $\mathcal{R}$-cycles have odd lengths.

SOLUTIONS
Proof. Let $a_0\mathcal{R}a_1\mathcal{R}a_2\mathcal{R}\dots a_n\mathcal{R}a_0$ be a cycle of the length $n+1$ and $a=a_0, b=a_1$. The points $a_i$ and $a_{i+1}$ lie on different sides of the line $l_{ab}$, $2 \le i \le n-1$, since any two big segments must intersect. Also, $a_n$ and $a_2$ should lie on the same side, otherwise the big segments $[a_n, a_0]$ and $[a_1, a_2]$ would not intersect. So, $n$ is even. $\square$
Lemma 3. There is an unique $\mathcal{R}$-cycle.
Proof. Let $a_0\mathcal{R}a_1\mathcal{R}a_2\mathcal{R}\dots a_n\mathcal{R}a_0$ be a cycle. Assume that there exists a big segment $[b, c]$ outside this cycle. The line of segment $[b, c]$ divides the plane into 2 semiplanes. Since big segments should intersect, adjacent (by $\mathcal{R}$) points of that cycle should lie on a different semiplanes. But it is impossible, because cycle contains an odd number of points. $\square$
Lemmas give a contradiction with assumptions, because $2006$ is even number.