More generally than the assertion of the problem, we prove that if we replace the number $2010$ in the problem by any even integer $2m$ bigger than or equal to $6$, we get for any convex $2m$-gon satisfying the condition of the problem the number $2m^2 - 4m + 1$ is the answer for the assertion.

For any integer $i$ less than or equal to $m$, let us agree to say that a diagonal of the polygon has the length $i$ if the diagonal connects two vertices which bound $i$ sides in between.

Let us first give an example of a Z.Z.L.S. for which there are exactly $2m^2 - 4m + 1$ self-intersection points. Let $P_1P_2 \cdots P_mQ_1Q_2 \cdots Q_m$ be the given $2m$-gon. For each $i = 1, 2, \dots, m-1$ connect vertices $P_i$ and $Q_{i+1}$, and vertices $Q_i$ and $P_{i+1}$. Each of these $2m-2$ diagonals has length $m-1$. Connect also the vertices $P_1$ and $Q_1$, and vertices $P_m$ and $Q_m$. These two diagonals have length $2$. It is easy to check that these $2m$ diagonals form a closed Z.Z.L.S.

Let us count the number of self-intersections of this closed Z.Z.L.S. In the sequel, the term "diagonal" refers only to diagonals constituting segments of this closed Z.Z.L.S.

Let us first show that there are $2m-4$ self-intersection points on the diagonal $P_iQ_{i+1}$. If we dissect the polygon into two parts by the diagonal $P_iQ_{i+1}$, then one part has $m$ vertices and the other $m-2$ vertices. There is no diagonal other than $P_{i+1}Q_i$ having length $m-1$ or more which is contained in either of these two parts. Consequently, the diagonal $P_iQ_{i+1}$ shares points with every one of the diagonals except $P_{i+1}Q_i$. So, it shares points with $2m-1$ diagonals. Of these $2m-1$ diagonals there are itself and two diagonals with which it shares an end point and these should be excluded from the consideration of self-intersection points, and therefore, there are $2m-4$ diagonals giving self-intersection points on the diagonal $P_iQ_{i+1}$. The same argument gives also that the diagonal $P_{i+1}Q_i$ has $2m-4$ self-intersection points as well.

For the diagonals $P_1Q_1$ and $P_mQ_m$ the same type of argument as above shows that both of them share points with all of the diagonals, and excluding itself and two diagonals which share an end point from the consideration, $2m-3$ diagonals give self-intersection points on each of these two diagonals.

Taking into account the fact that each self-intersection point is counted twice in this argument, we conclude, therefore, that there are
$$\frac{(2m-2)(2m-4) + 2 \cdot (2m-3)}{2} = 2m^2 - 4m + 1$$
self-intersection points for the closed Z.Z.L.S. constructed above.

Next, we show that there cannot be more than $2m^2 - 4m + 1$ self-intersections for any closed Z.Z.L.S.

By arguing as above, we can show that the number of self-intersection points on any diagonal cannot exceed $2m - 3$. Call a diagonal good diagonal if it contains $2m - 3$ self-intersection points. If there are $2$ or less good diagonals, then the total number of self-intersection points would be at most $2m^2 - 4m + 1$. So, we must have at least $3$ good diagonals.

Note that any good diagonal must have length $m$. To see why this is so, let us assume that a diagonal $l$ has length less than $m$. Then, when we dissect the polygon into two parts by $l$, one part has at most $m - 2$ vertices, and any diagonal intersecting with $l$ must have one of these vertices as its end point. Therefore, there are at most $2m - 4$ such diagonals, and $l$ cannot be a good diagonal.

Let us take three good diagonals. Since each one has length $m$, any pair of them intersect each other. So, we may designate them as $A_1A_4$, $A_2A_5$, $A_3A_6$, assuming that vertices $A_1$, $A_2$, $A_3$, $A_4$, $A_5$, $A_6$ lie on the polygon in this order.

Let $Q$ be a vertex of the polygon and let $Q'$ be the vertex lying at the other end of one of the diagonals starting from $Q$, and $Q''$ be the vertex lying at the other end of the diagonal starting from $Q'$ but different from $Q'Q$. If both $Q$ and $Q''$ are different from both $A_1$ and $A_4$ (then $Q'$ would be different from both $A_1$ and $A_4$ also), both diagonals $QQ'$, $Q'Q''$ intersect $A_1A_4$, since $A_1A_4$ is a good diagonal. Therefore, vertices $Q$ and $Q''$ lie on the same side of $A_1A_4$. The same reasoning applies to $A_2A_5$ and $A_3A_6$ as well, so we can conclude that there exists an integer $i$ ($1 \leq i \leq 6$) such that $Q$ and $Q''$ lie in between $A_i$ and $A_{i+1}$. Here we take $A_7$ to be equal to $A_1$, and we allow the possibility that $Q$ or $Q''$ may coincide with $A_i$ or $A_{i+1}$.

Let us color every other vertex on the closed Z.Z.L.S. black as we go along the cycle starting with a vertex. Then, for each of the pairs $A_1$, $A_4$, $A_2$, $A_5$, $A_3$, $A_6$, one and only one of the two vertices in the pair will be colored black. Without loss of generality we may assume that one of the following two cases occurs.

• $A_1$, $A_2$, $A_3$ are black.
• $A_1$, $A_3$, $A_5$ are black.

In the first case above, it should be possible to start from $A_1$ and repeating the process of going along $2$ contiguous diagonals to go through both $A_2$ and $A_3$ exactly once and return to $A_1$. But as we have seen above it would be impossible to go from $A_1$ to $A_3$ (and from $A_3$ to $A_1$) without going through $A_2$, so we have a contradiction.

In the second case, it is impossible to start from $A_1$ to go outside of the portion of the polygon bounded by $A_6$ and $A_2$ by repeating the process of going along $2$ contiguous diagonals, and hence it is impossible to reach from $A_1$ to $A_3$ by repeating the process of going along $2$ contiguous diagonals. Thus we have a contradiction in this case as well.

Summarizing, we have shown that there is no closed Z.Z.L.S. which has more than $2m^2 - 4m + 1$ self-intersections, and we have $2(1005)^2 - 4(1005) + 1 = 2016031$ as the solution to the problem.