jmc

Since $13$ is prime, all of the requested modular inverses exist. Furthermore, the inverses must be distinct: suppose that $a^{-1}\equiv b^{-1}(\mathrm{mod}13)$. Multiplying both sides of the congruence by $ab$, we obtain that $b\equiv ab\cdot a^{-1}\equiv ab\cdot b^{-1}\equiv a(\mathrm{mod}13)$.

Thus, the set of the inverses of the first $12$ positive integers is simply a permutation of the first $12$ positive integers. Then, $$\begin{array}{rl}& 1^{-1}+2^{-1}+\cdots +12^{-1}\\ & \equiv 1+2+\cdots +12\\ & \equiv 1+2+3+4+5+6\\ & +(-6)+(-5)+(-4)+(-3)+(-2)+(-1)\\ & \equiv \boxed{0}(\mathrm{mod}13).\end{array}$$