19. As shown in the figure, two ants, Jia and Yi, are moving on the following circle. $AC$ is the diameter of the large circle, and point $B$ is on $AC$. $AB$ and $BC$ are the diameters of the two smaller circles, respectively. Ant Jia crawls clockwise on the large circle, while Ant Yi crawls in the direction indicated by the arrows on the two smaller circles, forming a figure "8" $(A \rightarrow B \rightarrow C \rightarrow B \rightarrow A)$. Jia and Yi start from point $A$ at the same moment and continue to crawl. The speed ratio is $v_{\text {Jia }}: v_{\text {Yi}}=3: 2$. After $T_{1}$ minutes, the two ants meet. Then, Jia increases its speed by $\frac{1}{3}$, while Yi's speed remains unchanged, and they continue to crawl on their original paths. After $T_{2}$ minutes, the two ants meet again. It is known that $T_{1}+T_{2}=100^{3}-99^{3}+98^{3}-97^{3}+\cdots+2^{3}-1^{3}$, then the time it takes for Jia to crawl around the large circle at its original speed is $\qquad$ minutes (the answer should be written as an improper fraction).
【Analysis】The distance that Yi crawls for an 8-shaped path is $\pi \times A B+\pi \times B C=\pi \times A C$, and the distance that Jia crawls for one circle is $\pi \times A C$, so the distances crawled by Jia and Yi are equal, meaning the distance from $\mathrm{A}$ to $\mathrm{C}$ for Jia is the same as the distance from $\mathrm{A}$ to $\mathrm{C}$ for Yi.
Since initially $V_{\text {Jia }}: V_{\text {Yi }}=3: 2$, the first meeting occurs when Jia has crawled 3 circles and Yi has crawled 2 8-shaped paths, meeting at point $\mathrm{A}$.
After Jia increases his speed by $\frac{1}{3}$, $V_{\text {Jia }}^{*}: V_{\text {Yi }}=4: 2$, so the second meeting occurs when Jia has crawled 2 circles and Yi has crawled 1 8-shaped path.
$T_{1}+T_{2}$
$=(100-99)\left(100^{2}+100 \times 99+99^{2}\right)+\cdots+(2-1)\left(2^{2}+2 \times 1+1^{2}\right)$
$=100^{2}+99^{2}+\cdots+2^{2}+1^{2}+100 \times 99+\cdots+2 \times 1$
$=\frac{100 \times 101 \times 201}{6}+100^{2}-100+98^{2}-98+\cdots+2^{2}-2$
$=338350+2^{2} \times\left(50^{2}+49^{2}+\cdots+1^{2}\right)-(100+98+\cdots+2)$
$=338350+171700-2550$
$=507500$
Therefore, in 507500 minutes, Yi crawls 3 8-shaped paths, so the time it takes for Yi to crawl one 8-shaped path is $\frac{507500}{3}$ minutes.
Since initially $V_{\text {Jia }}: V_{\text {Yi }}=3: 2$,
Jia, at the initial speed, needs $\frac{507500}{3} \times \frac{2}{3}=\frac{1015000}{9}$ minutes to crawl one circle.