Let $ABC$ be a scalene triangle with inradius $1$ and exradii $r_A$, $r_B$, and $r_C$ such that \[20\left(r_B^2r_C^2+r_C^2r_A^2+r_A^2r_B^2\right)=19\left(r_Ar_Br_C\right)^2.\] If \[\tan\frac{A}{2}+\tan\frac{B}{2}+\tan\frac{C}{2}=2.019,\] then the area of $\triangle{ABC}$ can be expressed as $\frac{m}{n}$ for relatively prime positive integers $m$ and $n$. Compute $100m+n$.

[i]Proposed by Tristan Shin[/i]
1. We start with the given equation involving the exradii:
   \[
   20\left(r_B^2r_C^2 + r_C^2r_A^2 + r_A^2r_B^2\right) = 19\left(r_A r_B r_C\right)^2
   \]
   and the given sum of tangents:
   \[
   \tan\frac{A}{2} + \tan\frac{B}{2} + \tan\frac{C}{2} = 2.019
   \]

2. Recall that the inradius \( r \) of a triangle is related to the area \( \Delta \) and the semiperimeter \( s \) by:
   \[
   \Delta = sr
   \]
   Given \( r = 1 \), we have:
   \[
   \Delta = s
   \]

3. Using the identity for the tangents of half-angles in terms of the semiperimeter and sides:
   \[
   \tan\frac{A}{2} = \frac{r}{s-a}, \quad \tan\frac{B}{2} = \frac{r}{s-b}, \quad \tan\frac{C}{2} = \frac{r}{s-c}
   \]
   Since \( r = 1 \), we get:
   \[
   \frac{1}{s-a} + \frac{1}{s-b} + \frac{1}{s-c} = 2.019
   \]

4. Let \( x = s-a \), \( y = s-b \), and \( z = s-c \). Then:
   \[
   \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 2.019
   \]
   and:
   \[
   xy + yz + zx = 2.019xyz
   \]

5. From the given equation involving the exradii, we know:
   \[
   r_A = \frac{s}{s-a}, \quad r_B = \frac{s}{s-b}, \quad r_C = \frac{s}{s-c}
   \]
   Substituting these into the given equation:
   \[
   20\left(\left(\frac{s}{y}\right)^2 \left(\frac{s}{z}\right)^2 + \left(\frac{s}{z}\right)^2 \left(\frac{s}{x}\right)^2 + \left(\frac{s}{x}\right)^2 \left(\frac{s}{y}\right)^2\right) = 19\left(\frac{s}{x} \cdot \frac{s}{y} \cdot \frac{s}{z}\right)^2
   \]
   Simplifying, we get:
   \[
   20\left(\frac{s^4}{y^2z^2} + \frac{s^4}{z^2x^2} + \frac{s^4}{x^2y^2}\right) = 19\left(\frac{s^3}{xyz}\right)^2
   \]
   \[
   20\left(\frac{1}{y^2z^2} + \frac{1}{z^2x^2} + \frac{1}{x^2y^2}\right) = 19\left(\frac{1}{xyz}\right)^2
   \]

6. Let \( s = x + y + z \). Then:
   \[
   \frac{19s^2}{20} = x^2 + y^2 + z^2
   \]

7. Adding twice the product-sum identity:
   \[
   \frac{19(x+y+z)^2}{20} + 4.038xyz = x^2 + y^2 + z^2 + 2(xy + yz + zx)
   \]
   \[
   \frac{19(x+y+z)^2}{20} + 4.038xyz = (x+y+z)^2
   \]
   \[
   (x+y+z)^2 = 20 \cdot 4.038xyz
   \]

8. Using Heron's formula:
   \[
   \Delta = \sqrt{s(s-a)(s-b)(s-c)}
   \]
   Given \( \Delta = s \), we have:
   \[
   s = \sqrt{s(s-a)(s-b)(s-c)}
   \]
   \[
   s^2 = s(s-a)(s-b)(s-c)
   \]
   \[
   s = (s-a)(s-b)(s-c)
   \]

9. Substituting back:
   \[
   s^2 = 20 \cdot 4.038s
   \]
   \[
   s = 20 \cdot 4.038 = \frac{2019}{25}
   \]

10. The area of \( \triangle ABC \) is:
    \[
    \Delta = s = \frac{2019}{25}
    \]

The final answer is \( \boxed{ 100m + n = 201925 } \)