Can $\sum_1^n 1/k$ be arranged so that it is an integer for infinitely many $n?$ It's well known that when $n>1:$
$$\sum_{k=1}^n \frac{1}{k}\not\in \mathbb{N}$$
But if we are allowed to rearrange the series, we can for instance can get:
$$\frac{1}{1}\in\mathbb{N},\quad\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\in\mathbb{N},\quad \dots\;?$$ 
Maybe there is another obvious one that I didn't see.
Question: Can the harmonic series be rearranged so that its partial sums hit infinitely many integers?
 A: Answer : YES. It follows from the Erdös-Graham conjecture, which was proved in 2003 by Ernest E. Croot as indicated by Wikipedia.
I don’t know if there is a more direct proof. [EDIT : there is, as explained in Erick Wong’s comment below].
(If it is not obvious to you how it follows from the Erdös-Graham conjecture, consider a coloring of the integers that uses $n$ different colors, one for each integer between $2$ and $n$, and the last color for all integers $>n$. This shows that (1) there are Egyptian fractions summing to $1$ with arbitrarily large denominators. It follows that (2) For any integer $a>0$, there are Egyptian fractions summing to $a$ with arbitrarily large denominators, and hence that (3) for  any two integers $a,b>0$, there are Egyptian fractions summing to $\frac{a}{b}$ with arbitrarily large denominators). 
A: Yes, this is possible.
Start with $1=1/1$.  Now we've used $1$, so cross that off.  Then since $1$ is crossed off, write $1-1/2=1/2$; $1/2-1/3=1/6$; $1=1/2+1/3+1/6$, and cross off $2$, $3$, and $6$.  Then $1-1/4=3/4$ (since $1$, $2$, and $3$ are crossed off); $3/4-1/5=11/20$; $11/20-1/7=57/140$ (since $6$ is crossed off); etc.
So at each stage, we are trying to decompose $1$ into Egyptian fractions, but with certain fractions (those we've already used) forbidden.  We'll use the greedy algorithm to do this.  We need to show two things: Firstly, that each stage eventually terminates -- that we reach $1$, and move on to the next stage.  Secondly, that every Egyptian fraction gets used at some stage.
For the first of these, let us suppose that at a certain stage we have forbidden set $S$, and that at a certain step of that stage the number remaining left is $r/s$.  (Note that due to the setup, we're only ever dealing with fractions at most $1$, so $r\le s$.)  We're going to subtract off a fraction $1/k$.  Ordinarily, we'd have $k\le \frac{s}{r} + 1$, but since we have a forbidden set $S$, all we know is $k\le \frac{s}{r} + |S| + 1$.  That's still enough to imply that $1/k \ge \frac{r}{s+(|S|+1)r} \ge \frac{r}{(|S|+2)s}$; that is, we're slicing off at least a constant fraction of our fraction at each step, so it will go to zero (even if the process never terminates).
But that means that it will eventually be less than $1/M$, where $M$ is the largest element of $S$; and beyond that point, you are simply applying the greedy algorithm as normal, which we know yields a finite Egyptian fraction expansion.  This proves the first part.
The second part can be proven by strong induction.  Suppose we want to show that $1/n$ eventually gets used; assume that $1/m$ gets used for each $m<n$.  Then at the next stage after these are all used, $1/n$ will be the very first fraction used.
(Note, by the way, that since we're repeatedly decomposing $1$, we will not just hit infinitely many integers, we will hit every positive integer.)
By the way, I computed the first three stages of the process above; the numbers used are as follows:
First stage: 1
Second stage: 2, 3, 6
Third stage: 4, 5, 7, 8, 9, 10, 15, 230, 57960
