\soln
We will prove the following theorem.  The first two statements are 
special cases when $k=1$ and $k=2$.
\thm
For any integer $k\geq 1$, any subset of $[2kn]$ with $(2k-1)n+1$ elements
contains $2k$ numbers that sum to $2k^2n+k$.
\end{theorem}
\pf
We will use the strong pigeonhole principle.
Consider the following listing of the numbers 1 to $2kn$ in a
snake-like order.
\[
\begin{array}{c c c c c c c}
1 & 2 & 3 & \cdots & n-2 & n-1 & n \\
2n & 2n-1 & 2n-2 & \cdots & n+3 & n+2 & n+1 \\
2n+1 & 2n+2 & 2n+3 & \cdots & 3n-2 & 3n-1 & 3n \\
4n & 4n-1 & 4n-2 & \cdots & 3n+3 & 3n+2 & 3n+1 \\
4n+1 & 4n+2 & 4n+3 & \cdots & 5n-2& 5n-1 & 5n \\
6n & 6n-1 & 6n-2 & \cdots & 5n+3 & 5n+2 & 5n+1 \\
\vdots \\
(2k-2)n+1 & (2k-2)n+2 & (2k-2)n+3 & \cdots & (2k-1)n-2& (2k-1)n-1 & (2k-1)n \\
2kn & 2kn-1 & 2kn-2 & \cdots & (2k-1)n+3 & (2k-1)n+2 & (2k-1)n+1
\end{array}
\]
%\end{center}

The $(2k-1)n+1$ elements of $[2kn]$ are members of the preceding table.
There are $n$ columns, so by the pigeonhole principle, some column must have 
$2k$ elements.  
In other words, all the elements of one column must be present.
To complete the proof, we will show that each column sums to~$2k^2n+k$.

Notice that as we move one column to the right, the numbers
in the odd rows increase by one, and the numbers in
the even rows decrease by one.  Since there are an even number of
rows, every column has the same sum.  The sum of the first column is 
\begin{eqnarray*}
1+ \sum_{i=1}^{k-1} (2in + (2in + 1)) + 2kn & = &
1 + 4n\sum_{i=1}^{k-1} i + \sum_{i=1}^{k-1} 1 + 2kn \\
&=& 1 + 4n(k-1)k/2 + (k-1) + 2kn \\
&=& 2nk^2 + k.
\end{eqnarray*}
Therefore, the sum of each column is $2k^2n+k$.
\end{proof}