Given a set $S$ of $n$ positive integers $S=\{a_1,\ldots,a_n\}$, can we partition $S$ into $k$ subsets of equal sum such that each subset has contiguous elements from $S$?
Here, a contiguous subset is in the form $\{a_i,a_{i+1},a_{i+2},\ldots\}$ for some $i$. For example, with $S=\{3,6,2,1\}$, possible contiguous subsets are $\{3\}$, $\{3,6\}$, $\{3,6,2\}$, $\{6,2,1\}$, etc.
Is this variant of $k$-wat number partitioning NP-hard?
The problem is in $\mathsf{P}$ (and is not trivial), hence it is $\mathsf{NP}$-Hard if and only if $\mathsf{P} = \mathsf{NP}$.
In particular, you can design an easy linear-time greedy algorithm by observing that the sum of the numbers in each set of the partition must be exactly $T =\frac{1}{k} \sum_{i=1}^{n} a_n$. Since all $a_i$ are integers, the answer can only be yes if $T$ is an integer. Moreover, since all $a_i$ are positive, either there is no index $j$ such that the elements in $\{a_1, a_2,\dots, a_j\}$ sum to $T$, or such a $j$ is unique. In the latter case $\{a_1, a_2,\dots, a_j\}$ must be one of the sets of the partition, hence you can select it and repeat the above argument starting form $a_{j+1}$.
Here is the pseudocode of an iterative implementation of this algorithm, where a[i] represents $a_i$:
T = a[1] + a[2] + .... + a[n]
if T%n != 0:
return false
T = T/k //target sum of each set of the partition
sum = 0 //sum of the elements in the current set of the partition
for i = 1, ..., n:
sum = sum + a[i]
if sum > T: //no way to create the next set with sum T
return false
if sum = T: //we found the next set of the partition
sum = 0
return true
