The first equation shows us that $f(9+(x-9))=f(9-(x-9))\implies f(x)=f(18-x)$, while the second equation tells us that $f((x+10)-10)=(-(x+10)-10)\implies f(x)=f(-x-20)$. Therefore, $$f(x)=f(18-x)=f(x-38)$$This means, $f$ has a period of $38$. Therefore, $f(38n)=0$ for all integers $n$, and since $f(x)=f(18-x)$, this means $f(38n+18)=0$ for all integers $n$. Hence, $f$ has zeroes $0,38,76,\ldots,2014$ and $18,56,94,\ldots,1994$. This tells us that $f$ must have at least $54+53=\boxed{107}$ zeroes. To show that this minimum is achievable, just consider the following function: $$f(x)=\cos\frac{\pi(x-9)}{19}-\cos\frac{9\pi}{19}$$This function has zeroes precisely at where $x\equiv0\pmod{38}$ and $x\equiv18\pmod{38}$.