Find the number of $k$ satisfying $\frac{n}{3} - 1 < k \leq \frac{n}{2}$ for $n > 9$. Show that the number of $k \in \mathbb{N}$ satisfying $\frac{n}{3} - 1 < k \leq \frac{n}{2}$ for $n \in \mathbb{N}, n> 9$ is greater than $\frac{n}{6}$.
I looked at the distance $|\frac{n}{3} - 1 - (\frac{n}{2})| = 1 + \frac{n}{6}$ which is $> \frac{n}{6}.$ Is this correct or am I missing a proper justification? I'm not sure how to use the $n > 9$ fact either. Thank you.

Let $n = 6m + r$ for some non-negative integer $m$ and integer $0 \le r \le 5$. Thus, the question asks to prove there are more than $\frac{n}{6} = m + \frac{r}{6} \lt m + 1$ values of $k$, i.e., that there are at least $m + 1$ such values.
The inequality becomes
$$2m + \frac{r}{3} - 1 \lt k \le 3m + \frac{r}{2} \tag{1}\label{eq1}$$
In general for large divisors, you would normally look at various spans of values of $r$. However, since there aren't too many here, you can instead fairly easily manually check the results of the $6$ possible values of $r$ to get:
$$r = 0 \; \Rightarrow \; 2m - 1 \lt k \le 3m \tag{2}\label{eq2}$$
$$r = 1 \; \Rightarrow \; 2m - \frac{2}{3} \lt k \le 3m + \frac{1}{2}\tag{3}\label{eq3}$$
$$r = 2 \; \Rightarrow \; 2m - \frac{1}{3} \lt k \le 3m + 1 \tag{4}\label{eq4}$$
$$r = 3 \; \Rightarrow \; 2m \lt k \le 3m + \frac{3}{2} \tag{5}\label{eq5}$$
$$r = 4 \; \Rightarrow \; 2m + \frac{1}{3} \lt k \le 3m + 2 \tag{6}\label{eq6}$$
$$r = 5 \; \Rightarrow \; 2m + \frac{2}{3} \lt k \le 3m + \frac{5}{2} \tag{7}\label{eq7}$$
With \eqref{eq2} and \eqref{eq3}, $k$ goes from $2m$ to $3m$ inclusive, so there are $m + 1$ values which work. Similarly with \eqref{eq4}, $k$ goes from $2m$ to $3m + 1$, so there are actually $m + 2$ values which work. With \eqref{eq5}, $k$ goes from $2m + 1$ to $3m + 1$, for $m + 1$ values. Finally, \eqref{eq6} & \eqref{eq7} give $k$ going from $2m + 1$ to $3m + 2$, for $m + 2$ values. 
As such, this works for all $n \in \mathbb{N}$, including for $n \gt 9$. I'm not sure why this restriction is used in the question. Are there perhaps any other unstated conditions?
