If a particle travels 30 meters in any 3 seconds, does it necessarily travel 20 meters in any 2 seconds? Is it possible if its motion could've been discontinuous?
I'm trying to understand if there exists a function that has this property, but I chose to say it in terms of motion because it's easy to explain.

It isn't true, not even for continuous motion.
For example, let $V(t)=10+\cos\left(\dfrac{2\pi}{3}t\right)$ in meters/sec.
Then
\begin{eqnarray}
D&=&\int_0^3V(t)\,dt\\
&=&\left[10t+\frac{3}{2\pi}\sin\left(\dfrac{2\pi}{3}t\right)\right]_0^3\\
&=& 30
\end{eqnarray}
but
\begin{eqnarray}
D&=&\int_0^2V(t)\,dt\\
&=&\left[10t+\frac{3}{2\pi}\sin\left(\dfrac{2\pi}{3}t\right)\right]_0^2\\
&=& 20+\frac{3}{2\pi}\sin\left(\dfrac{4\pi}{3}\right)\\
&=&20-\frac{3\sqrt{3}}{4\pi}
\end{eqnarray}
For any three second time interval the distance traveled will be 30 meters because of the period of the sinusoidal portion of the velocity.
ADDENDUM: In response to a comment from @TheRubberDuck
\begin{eqnarray}
D&=&\int_x^{x+3}V(t)\,dt\\
&=&\left[10t+\frac{3}{2\pi}\sin\left(\dfrac{2\pi}{3}t\right)\right]_x^{x+3}\\
&=& 10(x+3)+\frac{3}{2\pi}\sin\left(\dfrac{2\pi}{3}(x+3)\right)-
\left(10x+\frac{3}{2\pi}\sin\left(\dfrac{2\pi}{3}x\right)\right)\\
&=&30+\frac{3}{2\pi}\left[ \sin\left(\frac{2\pi}{3}x+2\pi \right)-\sin\left(\dfrac{2\pi}{3}x\right)  \right]\\
&=&30
\end{eqnarray}
However, integrating between $x$ and $x+2$ gives a factor
\begin{eqnarray}
 \sin\left(\frac{2\pi}{3}x+\frac{4\pi}{3} \right)-\sin\left(\dfrac{2\pi}{3}x\right) 
\end{eqnarray}
which is not identically $0$.
