Given a fair coin and infinitely many tosses, must there exist some finite integer $k$ such that after you've done exactly $k$ total tosses, you'll get heads on tosses $k+1$ through $2k$?
This seems obviously true if you change the question to ask about tosses $k+1$ through $k+c$ for any constant $c$, but my guess is that once $k$ shows up as a multiplier, the situation changes. However, I'm not sure how to show this.
(Sorry if this is a repeat; I checked, but there are scads of these.)
For any integer $n$ the tosses $n+1,\ldots,2n$ are independent of the tosses $1,\ldots,n$, so as written clearly the statement cannot be true. I will assume that what you actually meant to ask was, "must there exist some finite integer $k$ such that you'll get heads on tosses $k+1$ through $2k$." In which case:
Let $0<p<1$ and $X_n$ a sequence of i.i.d. random variables with $\mathbb P(X_1=1)=p=1-\mathbb P(X_1=0)$. For each $n$, let $$ E_n = \bigcap_{j=n+1}^{2n} \{X_j = 1\}. $$ By independence, we have $$ \mathbb P(E_n) = \prod_{j=n+1}^{2n}\mathbb P(X_j = 1) = p^n. $$ Moreover, we have the bound $$ \mathbb P\left(\bigcup_{n=1}^\infty E_n\right) \leqslant \sum_{n=1}^\infty \mathbb P(E_n) = \sum_{n=1}^\infty p^n = \frac p{1-p}, $$ and in fact the inequality is strict as the $E_n$ are not disjoint. For $p=1/2$, it is clear that $\frac p{1-p}=1$, and so hence $$ \mathbb P\left(\bigcup_{n=1}^\infty E_n\right) < 1. $$ This means the statement is not true.