jmc

Since $16$ is even and only has a prime factor of $2$, all of the odd numbers are relatively prime with $16$ and their modular inverses exist. Furthermore, the inverses must be distinct: suppose that $a^{-1}\equiv b^{-1}(\mathrm{mod}16)$. Then, we can multiply both sides of the congruence by $ab$ to obtain that $b\equiv ab\cdot a^{-1}\equiv ab\cdot b^{-1}\equiv a(\mathrm{mod}16)$.

Also, the modular inverse of an odd integer $\mathrm{mod}16$ must also be odd: if the modular inverse of $m$ was of the form $2n$, then $2mn=16k+1$, but the left-hand side is even and the right-hand side is odd.

Thus, the set of the inverses of the first $8$ positive odd integers is simply a permutation of the first $8$ positive odd integers. Then, $$\begin{array}{rl}& 1^{-1}+3^{-1}+\cdots +15^{-1}\\ & \equiv 1+3+\cdots +15\\ & \equiv 1+3+5+7+(-7)+(-5)+(-3)+(-1)\\ & \equiv \boxed{0}(\mathrm{mod}16).\end{array}$$