TITLE: Cutting sticks puzzle
QUESTION [93 upvotes]: This was asked on sci.math ages ago, and never got a satisfactory answer.

Given a number of sticks of integral length  $ \ge n$ whose lengths
  add to $n(n+1)/2$. Can these always be broken (by cuts) into sticks of
  lengths $1,2,3, \ldots ,n$?

You are not allowed to glue sticks back together. Assume you have an accurate measuring device.
More formally, is the following conjecture true? (Taken from iwriteiam link below).

Cutting Sticks Conjecture: For all natural numbers $n$, and any given sequence $a_1, .., a_k$ of
  natural numbers greater or equal $n$ of which the sum equals
  $n(n+1)/2$, there exists a partitioning $(P_1, .., P_k)$ of $\{1, .., n\}$
  such that sum of the numbers in $P_i$ equals $a_i$, for all $1 \leq i \leq k$.

Some links which discuss this problem:

http://www.iwriteiam.nl/cutsticks.html

REPLY [8 votes]: I have implemented the suggestion I made in a comment to joriki's answer. For $3 \le n \le 18$, I have generated a list of subsets $S \subset \{1,2,...,n-1\}$ with the property that if a set of sticks with total length $n(n+1)/2$ takes all the lengths in $S$, together with any other lengths ≥n, then the sticks can always be cut into sticks of length $1,2,...,n$. It is available at this link (it's about 900K).  
I stared at it for a while, but nothing jumped out at me.
Edited to add: I have changed the program to output the sets in a more human-friendly order: part 1 (n = 1 to 17) and part 2 (n = 18).