jmc

The first equation is equivalent to the following: if $a+b=4$, then $f(a)=f(b)$. Similarly, the second equation is equivalent to the following: if $c+d=14$, then $f(c)=f(d)$.

Then note that for any $t$, we have $$f(t)=f(4-t)=f(t+10),$$because $t+(4-t)=4$ and $(4-t)+(t+10)=14$. This means that if $t$ is a root of $f$, then so is $t+10$, and conversely, if $t+10$ is a root of $f$, then so is $t$. Since $t=0$ is a root, we see that if $n$ is a multiple of $10$, then $f(n)=0$. We also have $f(4)=f(0)=0$, so if $n\equiv 4(\mathrm{mod}10)$, then $f(n)=0$.

To see that these are all the necessary roots, observe that $$f(x)=\{\begin{array}{rl}0& \text{if }x\text{ is an integer and either }x\equiv 0(\mathrm{mod}10)\text{ or }x\equiv 4(\mathrm{mod}10)\\ 1& \text{otherwise}\end{array}$$satisfies all the given conditions, and only has these roots. This is because if $a+b=4$ and $a\equiv 0(\mathrm{mod}10)$, then $b\equiv 4(\mathrm{mod}10)$, and vice versa. Similarly, if $c+d=14$ and $c\equiv 0(\mathrm{mod}10)$, then $d\equiv 4(\mathrm{mod}10)$, and vice versa.

There are $201$ multiples of $10$ in the given interval, and $200$ integers that are $4$ modulo $10$ in the given interval, making $201+200=\boxed{401}$ roots of $f.$