1. Write the general term

For $\left(x^{2/3}+\frac{1}{2}x^{-2/5}\right)^9,$ the general term is $T_{r+1}=\binom{9}{r}\left(x^{2/3}\right)^{9-r}\left(\frac{1}{2}x^{-2/5}\right)^r, \qquad r=0,1,2,\dots,9.$

So, $T_{r+1}=\binom{9}{r}\left(\frac12\right)^r x^{\frac{2}{3}(9-r)-\frac{2}{5}r}.$

2. Simplify the exponent of $x$

Exponent of $x$ in the general term: $\frac{2}{3}(9-r)-\frac{2}{5}r = 6-\frac{2r}{3}-\frac{2r}{5}.$

Taking LCM $15$, $6-\frac{10r+6r}{15}=6-\frac{16r}{15}.$

Thus, $T_{r+1}=\binom{9}{r}\left(\frac12\right)^r x^{6-\frac{16r}{15}}.$

3. Find coefficient of $x^{2/3}$

We need $6-\frac{16r}{15}=\frac23.$

So, $\frac{16r}{15}=6-\frac23=\frac{18-2}{3}=\frac{16}{3}.$

Hence, $r=\frac{15}{16}\cdot \frac{16}{3}=5.$

Therefore the coefficient of $x^{2/3}$ is $\binom{9}{5}\left(\frac12\right)^5=\frac{126}{32}=\frac{63}{16}.$

4. Find coefficient of $x^{-2/5}$

We need $6-\frac{16r}{15}=-\frac25.$

So, $\frac{16r}{15}=6+\frac25=\frac{30+2}{5}=\frac{32}{5}.$

Hence, $r=\frac{15}{16}\cdot \frac{32}{5}=6.$

Therefore the coefficient of $x^{-2/5}$ is $\binom{9}{6}\left(\frac12\right)^6=\frac{84}{64}=\frac{21}{16}.$

5. Add the two coefficients

$\frac{63}{16}+\frac{21}{16}=\frac{84}{16}=\frac{21}{4}.$

6. Check options

$\frac{21}{4}$ corresponds to Option D.

Therefore, the required sum is $\boxed{\frac{21}{4}}.$