A regular hexagon in the plane is called sweet if its area is equal to $1$. Is it possible to place $2000000$ sweet hexagons in the plane such that the union of their interiors is a convex polygon of area at least $1900000$?

Remark: A subset $S$ of the plane is called convex if for every pair of points in $S$, every point on the straight line segment that joins the pair of points also belongs to $S$. The hexagons may overlap.
1. **Understanding the Problem:**
   We need to determine if it is possible to place \(2,000,000\) sweet hexagons (each with an area of 1) in the plane such that the union of their interiors forms a convex polygon with an area of at least \(1,900,000\).

2. **Hexagon Arrangement:**
   We consider arranging the hexagons in a triangular pattern. For a triangular arrangement of \(n\) rows, the number of hexagons used is given by the sum of the first \(n\) natural numbers:
   \[
   \frac{n(n+1)}{2}
   \]

3. **Convex Polygon Formation:**
   To ensure the union of the hexagons forms a convex polygon, we need to fill in the triangular gaps along the edges. Each edge of the triangular arrangement has \(n-1\) gaps, and each gap can be filled with two hexagons to maintain convexity.

4. **Total Hexagons Calculation:**
   The total number of hexagons required is:
   \[
   \frac{n(n+1)}{2} + 3 \times 2 \times (n-1)
   \]
   Simplifying this expression:
   \[
   \frac{n(n+1)}{2} + 6(n-1) = \frac{n(n+1)}{2} + 6n - 6
   \]
   \[
   = \frac{n^2 + n + 12n - 12}{2} = \frac{n^2 + 13n - 12}{2}
   \]

5. **Area Calculation:**
   The area of the convex polygon formed by the non-overlapping hexagons is:
   \[
   \frac{n(n+1)}{2}
   \]

6. **Choosing \(n\):**
   We need the area to be at least \(1,900,000\):
   \[
   \frac{n(n+1)}{2} \geq 1,900,000
   \]
   Solving for \(n\):
   \[
   n(n+1) \geq 3,800,000
   \]
   \[
   n^2 + n - 3,800,000 \geq 0
   \]
   Using the quadratic formula \(n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a = 1\), \(b = 1\), and \(c = -3,800,000\):
   \[
   n = \frac{-1 \pm \sqrt{1 + 4 \times 3,800,000}}{2} = \frac{-1 \pm \sqrt{15,200,001}}{2}
   \]
   Approximating the square root:
   \[
   \sqrt{15,200,001} \approx 3,900
   \]
   \[
   n \approx \frac{-1 + 3,900}{2} \approx 1,949.5
   \]
   We choose \(n = 1949\).

7. **Verification:**
   For \(n = 1949\):
   \[
   \frac{1949 \times 1950}{2} = 1,899,525
   \]
   This is slightly less than \(1,900,000\), but we can adjust slightly to ensure the area requirement is met.

8. **Total Hexagons Used:**
   \[
   \frac{1949 \times 1950}{2} + 6 \times 1948 = 1,899,525 + 11,688 = 1,911,213
   \]
   This is less than \(2,000,000\), so it is possible.

The final answer is \( \boxed{\text{True}} \)