My solution: We say that at step $i$, we move $d_i$ units, where $d_i$ is an integer. From the condition we get that $|d_{i+1}-d{i}|$ is less than or equal to $1$ Suppose we fall two times on point $n$, there is a point to the right of $n$, or $n$ itself, such that the Flea makes a jump from it onto it (so it can go back). Let $m$ be this point. Take a coordinate system, where we will draw rectangles of base the segment between $i$ and $i+1$ and altitude $d_i$. Cleraly the sum (signed sum) of this rectangles from $0$ to $j$ is where the Flea is before the $j$ step. So suppose our Flea arrives to $m$, in $x$ steps for the last time, $d_x$ is $1$ at most, because the next step has distance $0$. so the Flea takes at most $x+1$ steps on steping back in $n$. Now draw the lines from $(0,1)$ with slope $1$, and from $(x,1)$ with slope $-1$, now traslade the altitude of the rectangles up, until they touch one of these lines with their leftmost vertex. The new area is bigger than or equal to the area of the rectangles, so in fact it is bigger than $m$ and $n$. But it is easy to calculate this area and from here obtain the bound we want for $x$

I solved it in a very similar way. If you want to pass through a point twice, you need to go there and go back (duh!). Since the length of the jumps stays the same or increases by $1$ or decreases by $1$, the flea must do a zero-jump (that is, a jump of lenght zero). In order to make it easier to see, represent each jump by a rectangle of width $1$ and height $d$, where $d$ is the length of the jump. Put these side by side, in order of the jumps. For example, if the flea made jumps of lengths $1$, $2$, $3$, $2$, $2$, $3$, $2$, $1$, $0$, $-1$, draw these rectangles. Each X is a unit square and the last X is a "negative square". X X XXXXXX XXXXXXXX X This is exactly the same interpretation Pascual did, and as he said before, the point where the flea gets is equal to the sum of the signed areas (or the number of unit squares) of the rectangles. For example, the area of the previous rectangles is $1 + 2 + 3 + 2 + 2 + 3 + 2 + 1 + 0 + (-1)$, not $1 + 2 + 3 + 2 + 2 + 3 + 2 + 1 + 0 + 1$. We are going to evaluate the further the flea can go with $k$ jumps, with the restriction that the flea is obliged to make a zero-jump. If $k = 2t$, the flea cannot make a jump bigger than $t$, since it takes $t+1$ jumps to make a $t+1$-lenght jump and more $t+1$ jumps to perform a zero jump. Thus the further the flea can go is to the point $1 + 2 + \cdots + t + (t-1) + (t-2) + \cdots + 1 + 0 = t^2$. Analogously, one can prove that if $k = 2t+1$, the further the flea can go is $t(t+1)$ (the same, but with another $t$-length jump). Thus if $n > t^2$, the flea must do more than $2t$ jumps, that is, it must perform at least $2t+1$ jumps; and if $n > t(t+1)$, it must do at least $2t+2$ jumps. It happens that if $t^2 < n \leq t(t+1)$ then $\lceil 2\sqrt n\rceil = 2t+1$ and it $t(t+1) < n \leq (t+1)^2$ then $\lceil 2\sqrt n\rceil = 2t+2$. Hence we are done.