jmc

If we first get a "common denominator" as if 3 and 13 represent real numbers rather than residues, we get $$\frac{1}{3}+\frac{1}{13}\equiv \frac{13+3}{39}\equiv \frac{16}{2\cdot 19+1}\equiv \frac{16}{1}\equiv \boxed{16}(\mathrm{mod}19).$$We can justify this as follows: let $a\equiv 3^{-1}(\mathrm{mod}19)$ and $b\equiv 13^{-1}(\mathrm{mod}19)$. Then $39a\equiv 13(\mathrm{mod}19)$ and $39b\equiv 3(\mathrm{mod}19)$. Summing these congruences shows that $39(a+b)\equiv a+b\equiv 13+3\equiv 16(\mathrm{mod}19)$, so $a+b\equiv 16(\mathrm{mod}19)$, as desired.