No there can be no such $E$.
Let $\chi_E$ be the characteristic function of $E$.
At every Lebesgue point $x$ of $\chi_E$, we have
$$\chi_E(x) = \lim_{r\to 0}\ \frac{1}{m(B_r)} \int_{B_r(x)} \chi_E(y) \ dy$$
Since almost every point of an integrable function is a Lebesgue point, the set of $x$ for which this limit is not equal to $0$ or $1$ must have measure $0$.
In particular, it cannot be true that
$$m(E\cap[a,b]) = \frac{b-a}{2}$$
for all $a,b$, since that would imply that also
$$\frac{1}{2} = \frac{m(E\cap[x-r,x+r])}{m([x-r,x+r])} = \frac{1}{m(B_r)} \int_{B_r(x)} \chi_E(y) \ dy$$
for all $x$ and all small enough $r>0$. (Which would be in contradiction to the fact about Lebesgue points above.)