Ukrainian National Mathematical Olympiad

Two parabolas with different leading coefficients are tangent if and only if they have exactly one common point, so the equation $x^2+2013x+a=-x^2+ax+2013$, that is equivalent to the equation $2x^2+(2013-a)x+(a-2013)=0$, must have exactly one root. Therefore, $$D=(a-2013{)}^2-8(a-2013)=(a-2013)(a-2021)=0.\text{ Hence, }a=2013,\text{ or }a=2021.$$