Suppose that $l$ is a line in $L$. Suppose that $p_1,p_2,\ldots,p_{2005}$ are the intersection points of other lines with $l$. Also suppose that $p_i$s are arranged in order in some direction on the line $l$. Then exactly the lines which have the points $p_1$ and $p_{2005}$ are bounding the line $l$. For the line $l$ let $s(l)=p_1p_{2005}$ be the segment between $p_1$ and $p_{2005}$. Suppose the wanted result is not correct, so if $l$ bounds $l'$ then $l'$ bounds $l$. Construct a graph with $L$ as its vertex set. Draw an edge between two lines if one bounds the other. In this graph degree of every vertex is $2$ so the graph is composed of some cycles. Suppose that $l_1,l_2,\ldots,l_k$ is a cycle. Suppose that $P_i$ is the comon endpoint of $s(l_i)$ and $s(l_{i-1})$ ($l_{-1}=l_k$). If $k$ is even, then $P_3P_4$ intersects $l_1$ so $P_4$ and $P_3$ are on two sides of $l_1$. $P_4P_5$ intersects $l$, so $P_5$ and $P_3$ are on the same side of $l_1$. Using induction we see that $P_{odd}$ and $P_3$ are on the same side of $l_1$ and $P_{even}$ is on the other side. So $P_k$ and $P_3$ are on opposite sides of $l_1$ so $P_2P_3$ and $P_kP_1$ can not intersect which is contradiction. If $k$ is odd then $k<2006$ so there is another line like $l$. Now using similar arguments we see that $P_{odd}$s and $P_1$ are on the same side of $l$ and $P_{even}$s are on the other side, so $P_k$ and $P_1$ are on the same side and so $P1P_k$ can not intersect $l$ which is also contradiction. So we always arrive at a contradiction, so the problem is proved!