On extending a function $f:\mathbb N \to [-1,1]$ to $\bar f:\mathbb R \to [-1,1]$ such that $\bar f$ is differentiable at every point of $\mathbb N$? For which type of functions $f:\mathbb N \to [-1,1]$ , can we extend it to a continuous function $\bar f:\mathbb R \to [-1,1]$ such that $\bar f$ is differentiable at every point of $\mathbb N$ ? 
And for which type of functions $f:\mathbb N \to [-1,1]$ , can we extend it to a function $\bar f:\mathbb R \to [-1,1]$ such that $\bar f$ is differentiable at every point of $\mathbb N$ ( this time , not requiring $\bar f$ to be continuous everywhere )  ? 
By extension of $f$ , I obviously do mean $f(\mathbb N)=\bar f(\mathbb N)$
 A: We can do the first paragraph easily.  Just use a constant $\overline f(x)$ of length $\frac 12$ around each point of $\Bbb N$ and a straight line over he intervening intervals to connect them.  
We can do much better than you are asking.  We can extend all of them so $\overline f(x)$ is differentiable at every point of $\Bbb R$  Let us focus on he interval between $1$ and $2$. If $f(1)=f(2)$, join them with a straight line.  If $f(1)=1, f(2)=-1$, let $$\overline f(x)=\begin {cases} 1&1\le x \lt m\\1-k(x-m)^2&m \le x \lt \frac 32\\-1+k(x-(3-m))^2&\frac 32\le x \lt3-m\\-1&3-m \le x \lt 2\end {cases}$$  You should be able to find $k,m$ so this works with a derivative at $\frac 32$.  The point is that $m$ and $3-m$ are symmetric in the interval.  You have two equations in two unknowns, one from requiring that $\overline f(\frac 32)=0$ from both sides and one from requiring that the derivatives match at $\frac 32$  
We can do much better than that.  We can make a $C^\infty$ function that goes through all your points.  Look up a bump function and make a bump that goes through all the defined points and returns to $0$ before the next one.
A: Using polynomial spline, one can see that any sequence $\{f(n)\}_{n=1}^\infty$ can be extended in this manner.
