1729 is interesting!

First we have to prove that $I(17;29)$ is the ellipse center. Suppose $P(x_0;y_0)$ be a point on the ellipse $(E)$ . That means: $\frac{(x_0-17)^2}{17}+\frac{(y_0-29)^2}{29}=1729$ $\implies$ $\frac{((34-x_0)-17)^2}{17}+\frac{((58-y_0)-29)^2}{29}=1729$ $\implies$ $P'(34-x_0;58-y_0)$ is also on $E$ . Futhermore, $\overrightarrow{IP}=(x_0-17;y_0-29)=-(17-x_0;29-y_0)=-\overrightarrow{IP'}$ . Therefore, $I$ is the center of $(E)$ .

Now, as the center is found, we have another diagram of $(E)$ .

Suppose S is the area of $(E)$ , $A+B+C+D+E$ is the area of the colored regions. Based on the graph, we have:

$S_1=\frac{S}{4}+A+C+E$

$S_2=\frac{S}{4}-A+B$

$S_3=\frac{S}{4}-B-C-D$

$S_4=\frac{S}{4}-E+D$

$\implies$ $M=S_1+S_3-S_2-S_4=\frac{S}{4}+A+C+E+\frac{S}{4}-B-C-D-\frac{S}{4}+A-B-\frac{S}{4}+E-D=2A+2E-2B-2D$

$A=C+D;E=B+C$ so $M=4C=4.17.29=\boxed{1972}$ .

1972 is my mom's birthyear and today is her birthday. This problem is dedicated to her.