Let $K$ be the set of all sides and diagonals of a convex $2010-gon$ in the plane. For a subset $A$ of $K,$ if every pair of line segments belonging to $A$ intersect, then we call $A$ as an [i]intersecting set.[/i] Find the maximum possible number of elements of union of two [i]intersecting sets.[/i]
1. **Labeling and Initial Setup:**
   - Label the vertices of the convex $2010$-gon as $1, 2, \ldots, 2010$ in clockwise order.
   - Consider the polygon to be regular for simplicity.

2. **Constructing an Intersecting Set:**
   - We can construct an intersecting set by considering diagonals and sides that intersect at a common vertex.
   - For example, consider the set $\{(1, 3)\} \cup \{(2, i) \mid i \neq 2\}$ and $\{(15, 17)\} \cup \{(16, i) \mid i \neq 16\}$.

3. **Counting Elements:**
   - Each of these sets contains $2010$ elements (one for each vertex except the starting vertex).
   - The union of these two sets will have $4019$ elements since they share one common element.

4. **Proving Optimality:**
   - **Lemma:** Any intersecting set has at most $2010$ diagonals.
     - **Proof:** Define the value of a diagonal connecting vertices $i, j$ to be the unique integer $k$ in $[0, 2010)$ such that $i + j \equiv k \pmod{2010}$.
     - No two diagonals with the same value can intersect, as they would be parallel.
     - Therefore, all diagonals in an intersecting set must have distinct values, implying the set can have at most $2010$ elements. $\blacksquare$

5. **Disjoint Intersecting Sets:**
   - Suppose, for contradiction, that there exist two disjoint intersecting sets $A$ and $B$ each of size $2010$.
   - By the lemma, each set contains exactly one diagonal of every value.
   - Let $XY \in A$ and $ZT \in B$ be diagonals of value $0$.
   - Assume $X, Y, Z, T$ lie on the circumcircle of the polygon in clockwise order.
   - At time $t = 0$, $d_1 = XY$ and $d_2 = ZT$.
   - At time $t = i$, change $d_1, d_2$ to the unique diagonals in $A, B$ respectively with value $i$.
   - Since $A$ and $B$ are intersecting sets, $d_1$ and $d_2$ pivot clockwise about one of their endpoints every time.
   - $d_1$ must always have at least one endpoint in the arc $XY$ not containing $ZT$, and similarly for $d_2$.
   - These observations imply that $d_1$ and $d_2$ must eventually coincide, leading to a contradiction.

6. **Conclusion:**
   - Since $4020$ diagonals are unattainable, the maximum possible number of elements in the union of two intersecting sets is $4019$.

The final answer is $\boxed{4019}$