Question:
What is the residue modulo $16$ of the sum of the modulo $16$ inverses of the first $8$ positive odd integers?

Express your answer as an integer from $0$ to $15$, inclusive.

Answer:
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 $a^{-1} \equiv b^{-1} \pmod{16}$. Then, we can multiply both sides of the congruence by $ab$ to obtain that $b \equiv ab \cdot a^{-1} \equiv ab \cdot b^{-1} \equiv a \pmod{16}$.

Also, the modular inverse of an odd integer $\mod{16}$ 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{align*}&1^{-1} + 3^{-1} + \cdots + 15^{-1} \\
&\equiv 1 + 3 + \cdots + 15 \\ &\equiv 1 + 3 + 5 + 7 + (-7) + (-5) + (-3) + (-1) \\ &\equiv \boxed{0} \pmod{16}.\end{align*}