$k$ is an odd number, $(n,k!)=1$, prove that $$\sum_{i=k}^{n-1}\left\{\frac{\binom{i}{k}}{n}\right\}=\frac{n-k^{w(n)}}{2},$$ where $\{x\}=x-[x]$, $w(n)$ is the number of distinct prime factors of $n$.
For example, if $p>k$ is prime then $$\sum_{i=k}^{p-1}\left\{\frac{\binom{i}{k}}{p}\right\}=\frac{p-k}{2}.$$
When $k=1,$ the proof is trivial. Suppose $k\geq 3$ and $n$ has one distinct prime factor (meaning $n$ is a prime greater than $k$). Since both $n$ and $k$ are odd, the number of terms $n-k$ in the sum is even.
Define $C_i = \binom{i}{k}.$ Consider the sum $C_{k+j} + C_{n-1-j}$ for $j=0\dots (n-k)/2-1:$ $$ \begin{align} \binom{k+j}{k} + \binom{n-1-j}{k} &= \frac{(k+j)!}{k! j!} + \frac{(n-1-j)!}{k! (n-1-j-k)!} \nonumber \\ &= \frac{(k+j)!(n-1-j-k)! + (n-1-j)!j!}{k! j! (n-1-j-k)!}. \tag{1}\label{eqn1} \end{align} $$ If this sum is divisible by $n,$ the sum of the the remainders when dividing $C_{k+j}$ and $C_{n-1-j}$ by $n$ must be $n$ (since neither $C_{k+j}$ nor $C_{n-1-j}$ is divisible by $n$).
The numerator of \eqref{eqn1} can be written $$ (n-(j+k+1))! \big((k+j)! + j!(n-(j+k))(n-(j+k-1))\cdots (n- (j+1))\big). $$ The term $$ j!(n-(j+k))(n-(j+k-1))\cdots (n-(j+1)) $$ is a polynomial in $n$ with constant term $-(j+k)!,$ so the numerator of \eqref{eqn1} is divisible by $n.$ So the sum of all the remainders of the $C_i$ when divided by $n$ is $n(n-k)/2;$ dividing the sum of remainders by $n$ gives the sum in the problem.
If $n$ has multiple distinct prime factors, each factor must be greater than $k$ and $n$ must still be odd. I have not completely worked out the proof in this case, but I don't think it's too hard. It boils down to computing the number of pairs of remainders that sum to zero rather than $n$ because $C_{k+j}$ and $C_{n-1-j}$ are divisible by all the prime factors of $n.$