For every positive integer $k$, let $\mathbf{T}_k = (k(k+1), 0)$, and define $\mathcal{H}_k$ as the homothety centered at $\mathbf{T}_k$ with ratio $\tfrac{1}{2}$ if $k$ is odd and $\tfrac{2}{3}$ is $k$ is even. Suppose $P = (x,y)$ is a point such that 
$$(\mathcal{H}_{4} \circ \mathcal{H}_{3} \circ \mathcal{H}_2 \circ \mathcal{H}_1)(P) = (20, 20).$$ What is $x+y$?
(A [i]homothety[/i] $\mathcal{H}$ with nonzero ratio $r$ centered at a point $P$ maps each point $X$ to the point $Y$ on ray $\overrightarrow{PX}$ such that $PY = rPX$.)
1. Start with the point \( P = (x, y) \).

2. Apply the homothety \(\mathcal{H}_1\) with ratio \(\frac{1}{2}\) centered at \(\mathbf{T}_1 = (2, 0)\):
   \[
   \mathcal{H}_1(P) = \left( \frac{x + 2}{2}, \frac{y}{2} \right)
   \]

3. Apply the homothety \(\mathcal{H}_2\) with ratio \(\frac{2}{3}\) centered at \(\mathbf{T}_2 = (6, 0)\):
   \[
   \mathcal{H}_2 \left( \mathcal{H}_1(P) \right) = \mathcal{H}_2 \left( \frac{x + 2}{2}, \frac{y}{2} \right) = \left( \frac{\frac{x + 2}{2} + 6}{\frac{3}{2}}, \frac{\frac{y}{2}}{\frac{3}{2}} \right) = \left( \frac{x + 8}{3}, \frac{y}{3} \right)
   \]

4. Apply the homothety \(\mathcal{H}_3\) with ratio \(\frac{1}{2}\) centered at \(\mathbf{T}_3 = (12, 0)\):
   \[
   \mathcal{H}_3 \left( \mathcal{H}_2 \left( \mathcal{H}_1(P) \right) \right) = \mathcal{H}_3 \left( \frac{x + 8}{3}, \frac{y}{3} \right) = \left( \frac{\frac{x + 8}{3} + 12}{2}, \frac{\frac{y}{3}}{2} \right) = \left( \frac{x + 44}{6}, \frac{y}{6} \right)
   \]

5. Apply the homothety \(\mathcal{H}_4\) with ratio \(\frac{2}{3}\) centered at \(\mathbf{T}_4 = (20, 0)\):
   \[
   \mathcal{H}_4 \left( \mathcal{H}_3 \left( \mathcal{H}_2 \left( \mathcal{H}_1(P) \right) \right) \right) = \mathcal{H}_4 \left( \frac{x + 44}{6}, \frac{y}{6} \right) = \left( \frac{\frac{x + 44}{6} + 20}{\frac{3}{2}}, \frac{\frac{y}{6}}{\frac{3}{2}} \right) = \left( \frac{x + 104}{9}, \frac{y}{9} \right)
   \]

6. Given that the final point is \((20, 20)\), we set up the equations:
   \[
   \frac{x + 104}{9} = 20 \quad \text{and} \quad \frac{y}{9} = 20
   \]

7. Solve for \(x\) and \(y\):
   \[
   \frac{y}{9} = 20 \implies y = 180
   \]
   \[
   \frac{x + 104}{9} = 20 \implies x + 104 = 180 \implies x = 76
   \]

8. Therefore, \(x + y = 76 + 180 = 256\).

The final answer is \(\boxed{256}\)