I am looking at all sequences $\lbrace a_i\rbrace_{i=1}^{2n}$ so that $ a_i = \pm 1$ and $\sum_{i=1}^{2n} a_i =0$. Let me call such a sequence "balanced" (of length $2n$).
Question: For which $k$ does the following hold: that for any balanced sequence $\lbrace a_i\rbrace_{i=1}^{2n}$, there exists $j$ so that $\lbrace a_i\rbrace_{i=j}^{j+2k-1}$ is balanced (of length $2k$).
I've managed to show this $k$ divides $n$, but there are lots of extra casses where it works. Indeed, for $n \leq 11$ the exact condition is $k \leq \frac{n+1}{2}$.
However, starting at $n=12$ this is no longer true. Look at the sequence $++++------++++------++++$. There is no sequence of length 10 (so $k=5$) which is balanced.
Furthermore, you can construct counterexamples where $k$ is really small in comparison to $n$. Here is a family of counterexamples. Assume $n = ja = (j+1)b$ where $a$ and $b$ are even and $j$ is some integer. Consider the sequence $b$ "-", $a$ "+", $b$ "-", $a$ "+", .... , $b$ "-", $a$ "+" and $b$ "-". Then any sequence with $k = \tfrac{1}{2}(a+b)$ will contain $a$ "+" and $b$ "-" (which is not balanced).