jmc

Because $f(x)$ has real coefficients and $2i$ and $2+i$ are zeros, so are their conjugates $-2i$ and $2-i$. Therefore

$$\begin{array}{rl}f(x)=(x+2i)(x-2i)(x-(2+i))(x-(2-i))& =(x^2+4)(x^2-4x+5)\\ & =x^4-4x^3+9x^2-16x+20.\end{array}$$Hence $a+b+c+d=-4+9-16+20=\boxed{9}$.