Positioning a member of an interval partition Let $\ 0<\Lambda_1\le\ldots\le\Lambda_n\ $ be a finite non-decreasing sequence of positive reals, of length $\ n>0.\ $ Let
$$ D:=\sum_{k=1}^n \Lambda_k $$
The question is about the conditions which
allows to shift the interval $\ [0;\Lambda_k)\ $ over arbitrary
$\ x\in[0;D)\ $ by applying a permutation
$\ \pi:\{1\ldots n\}\rightarrow\{1\ldots n\}\ $ to the members
$\ [0;\Lambda_k)\ $ of partition $\ [0;D).\ $ The following simple theorem will provide the picture, then I will reformulate my question in a precise way.

Notation: $\ \mathbb S(n)\ $ is the symmetric group (of all permutations of $\ \{1\cdots n\}),\ $ and the union of the $n$
  half_closed-half_open intervals (for $k=1\ldots n$)
$$ L(\pi\ k)\ :=\ \left[\sum_{t=1}^{\pi(k)-1}\Lambda_{\pi(t)};
                          \sum_{t=1}^{\pi(k)}\Lambda_{\pi(t)} \right)$$
  partitions $\ [0;D)\ $ for every $\ \pi\in\mathbb S(n)$.

THEOREM 1
$$ [0;D)\ =\ \bigcup_{\pi\in\mathbb S(n)} L(\pi\ n)$$

Remark: It is essential to have here intervals for $\ k=n.\ $
  In general, the above theorem doesn't hold for $\ k<n,\ $ i.e. when
  $\ \Lambda_k\ <\ \max_{t=1\ldots n} \Lambda_t$.

QUESTION   What is the necessary and/or sufficient condition for
$$ [0;D)\ =\ \bigcup_{\pi\in\mathbb S(n)} L(\pi\ k)$$
This question is still a shorthand for several related questions, be it 
about a single index $k$ or about a group of them.
Example 1   If $\ \Lambda_n>\frac D2\ $ then the answer to the Question is NO for every $ k<n;\ $ indeed, for such $k,\ $ no permutation can shift $\ [0;\Lambda_k)\ $ over $\ \frac D2$.
More generally,
if
$\ 2\cdot s<n\ $ and $\ \sum_{t=n-2\cdot s}^{n-s}\Lambda_t\ >\ \frac D2\ $
then the answer to Question is NO for every $k<n-2\!\cdot\! s$.


Backround The above theorem 1 was motivated and serves me as a step (tiny but essential) in my study of Egyptian sums.

 A: At the other end of the above statements, if $\ n>1\ $ and
$\ \Lambda_1<\Lambda_2\ $ then the answer is NO for $\ k=1:$
THEOREM 2   Let $\ n>1.\ $ Then
$$ \bigcup_{\pi\in\mathbb S(n)} L(\pi\ 1)\,\ \subseteq
   \,\ [0;\Lambda_1)\,\cup\,[\Lambda_2;D-\Lambda_2)\,\cup\,
    [\Lambda_n-\Lambda_1;D) $$
We see that $\,\ [0;D)\,\ne\,\bigcup_{\pi\in\mathbb S(n)} L(\pi\ 1)\,\ $
when $\ \Lambda_1<\Lambda_2\ $ since the non-empty intervals
$\ [\Lambda_1;\Lambda_2)\ $ and
$\ [D-\Lambda_n-\Lambda_2;\ D-\Lambda_n-\Lambda_1)\ $ are totally
ommitted.
More generally,
THEOREM 3
$$ \bigcup_{\pi\in\mathbb S(n)} L(\pi\ r)\,\ \subseteq
   \,\ \left[0;\sum_{k=1}^s\Lambda_k\right)\ \cup 
   \ [\Lambda_{s+1};\ D-\Lambda_{s+1})\ \cup
   \ \left[D-\sum_{k=1}^s\Lambda_k;\ D\right) $$
where $\ r\le s<n$. Thus, the answer is NO for every $\ r\le s\ $ whenever
$$ \sum_{k=1}^s\Lambda_k\ <\ \Lambda_{s+1} $$
And a positive theorem follows,
THEOREM 4   If $\ \Lambda_n\le\frac D2\ $ and $\ n>1\ $ then the answer is  YES  for $\ n-1,\ $ and moreover
$$ \bigcup_{\pi\in\mathbb S(n)} L(\pi\,\ n\!-\!1)\,
   \ =\,\ [0;D)\ \quad\Leftarrow:\Rightarrow\quad\ \Lambda_n\,\le\,\frac D2 $$
Finally, more generally (once again),
THEOREM 5 If $\ 1\le s < n,\ $ and
$\ \sum_{k=1}^s\Lambda_k\ge\frac D2\,\ $ then the answer is  YES 
for every $\ u=s\ldots n,\ $ i.e.
$$  \bigcup_{\pi\in\mathbb S(n)} L(\pi\,\ u)\,
   \ =\,\ [0;D) $$
