If all you care about is the valid numbers of attendants, and not the many possible ways to realize the splitting, you can do it as follows: for each $k$, the number of attendants can be anywhere from $kn-2k=k(n-2)$ through $kn+2k=k(n+2)$. To see this, note that you can achieve $k(n-2)$ by dividing them into $k$ groups of $n-2$ each; you can achieve $k(n-2)+1$ by having one group of $n-1$ and the rest of $n-2$; you can achieve $k(n-2)+2$ by having two groups of $n-1$ and $k-2$ of $n-2$; and so on until you get to $kn-k$, which can be achieved with $k$ groups of $n-1$. Then you can add one person to each group one at a time to cover $kn-k$, $kn-k+1,\ldots,kn$. Then add one person to each group to get an instance for $kn+1$ through $kn+k$; and yet again to get all the ones from $kn+k+1$ through $kn+2k$.
Moreover, if the number of attendants is strictly between $k(n+2)$ and $(k+1)(n - 2)$ for some value of $k$, then achievement is impossible: having at most $k$ groups is insufficient (maximum number of attendants is $kn+2k$), and having $k+1$ groups is too many (minimum number of attendants is $(k+1)(n-2)$). So no number strictly between $k(n+2)$ and $(k+1)(n-2)$ can be achieved.
Now, there comes a point after which $k(n+2)\geq (k+1)(n-2)$; namely, $\begin{align*} k(n+2) &\geq (k+1)(n-2)\\ k(n+2) &\geq k(n-2) + n-2\\ k(n-2) + 4k &\geq k(n-2) + n-2\\ 4k &\geq n-2\\ 4k+2 &\geq n. \end{align*}$ So, if $k$ is the smallest integer such that $4k+2\lt n$ (namely, $k=\lceil \frac{n-2}{4}\rceil$), then there are no unachievable numbers greater than $k(n-2)$, because there will be no "gap" between the case of $k$ groups and the case of $k+1$ groups.
Added. In fact, this can be strengthened a bit, as Henry points out. It's enough for $k(n+2)$ to be exactly one less than $(k+1)(n-2)$ for there to be no unachievable numbers. Solving for that, we have: $\begin{align*} k(n+2)+1 &\geq (k+1)(n-2)\\ k(n-2) + 4k+1 &\geq k(n-2) + n-2\\ 4k+1 &\geq n-2\\ 4k+3 &\geq n. \end{align*}$ So it is enough for $k\geq \frac{n-3}{4}$, or $k=\lceil\frac{n-3}{4}\rceil$ to ensure that there are no gaps after $k(n-2)$.
So if $\ell = \lceil \frac{n-3}{4}\rceil$, then the achievable numbers are:
$n-2$, $n-1$, $n$, $n+1$, $n+2$, $2n-4$, $2n-3$, $2n-2$, $2n-1$, $2n$, $2n+1$, $2n+2$, $2n+3$, $2n+4$, $3n-6,\ldots,3n+6$, $4n-8,\ldots,4n+8,\ldots$, $(\ell -1)(n-2)$, $(\ell-1)(n-2)+1,\ldots,(\ell-1)(n+2)$, and all integers greater than or equal to $\ell(n-2)$.