A k-approximation to k-way number partitioning
This is not an answer - it is just a reply to Neal Young's comment regarding the greedy algorithm. I claim that in the outcome of the greedy algorithm:

The largest sum is at most $\operatorname{MinMax}(S, \lfloor(k+1)/2\rfloor)$;
The smallest sum is at least $\operatorname{MaxMin}(S, 2k-1)$.

Proof. Let $X_1,\ldots,X_k$ be the outcome of the greedy algorithm. It is sufficient to prove the claims for an arbitrary subset from this partition, say for $X_1$. For any set $X$, denote by $v(X)$ the sum of numbers in $X$.
Proof of 1.
Let  $Z_1,\ldots,Z_{\lfloor(k+1)/2\rfloor}$ be a min-max partition.
If $X_1$ contains a single item, then it is contained in at least one $Z_i$, so $v(X_1)\leq \max_i v(Z_i)$.
Otherwise, let $x_1$ be the last item added to $X_1$. By the greedy algorithm definition, both $v(x_1)$ and $v(X_1\setminus \{x_1\})$ are smaller than $v(X_2),\ldots,v(X_k)$. So in the $(k+1)$-partition $\{x_1\},X_1\setminus \{x_1\},X_2,\ldots,X_k$, the first two elements are smallest. Therefore, their sum is at most $2/(k+1)$ of the total:
\begin{align*}
v(X_1) \leq \frac{2}{k+1}v(S)
\end{align*}
On the other hand:
\begin{align*}
\max_i v(Z_i) \geq \frac{1}{\lfloor(k+1)/2\rfloor}v(S)
\geq
\frac{2}{k+1}v(S)
\end{align*}
Therefore, $v(X_1)\leq \max_i v(Z_i)$ too.
Proof of 2.
Let $x_2,\ldots,x_k$ be the last item added to $X_2,\ldots,X_k$ respectively.
By the greedy algorithm definition, if we remove $x_2,\ldots,x_k$ from $X_2,\ldots,X_k$, then $v(X_1)$ is largest among the remaining subsets. Therefore,
\begin{align*}
v(X_1)\geq \frac{1}{k}v(S\setminus \{x_2,\ldots,x_k\})
\end{align*}
Let $Y_1,\ldots, Y_{2k-1}$ be an optimal $(2k-1)$ partition.
The items $x_i$ belong to at most $n-1$ sets in the partition $Y$, so there are at last $n$ sets in that partition that do not contain any $x_i$; denote them by $Y_1,\ldots,Y_k$. So
\begin{align*}
\min_{i=1}^{2k-1} v(Y_i) 
\leq 
\min_{i=1}^{k} v(Y_i) 
\leq
\frac{1}{k}v(Y_1\cup\cdots \cup Y_k)
\leq
\frac{1}{k}v(S\setminus \{x_2,\ldots,x_k\})
\end{align*}
Therefore, $v(X_1)\geq \min_i v(Y_i)$.
So the greedy algorithm is a 2-approximation. The question of whether there exists an additive approximation remains open.