**Part I: $CN^{1/3}$ is enough.**

Start with $P(x)=\prod_{k\le N^{1/3}}(1-k^2x^2)$. Notice that it vanishes at $1/k$ with $k\le N^{1/3}$, is bounded by $1$ on $[0,N^{-1/3}]$, the degree of $P$ is $2N^{1/3}$ and on the interval $[0,2N^{-1}]$ we have $\log P\ge -2\sum_{k\le N^{1/3}}\frac{4k^2}{N^2}=o(1)$. Now just take the Chebyshev polynomial $Q$ of degree $N^{1/3}$ adjusted to the interval $[0,N^{-1/3}]$. It will jump as you requested between $N^{-1}$ and $2N^{-1}$ and stay below $1$ on $[0,N^{-1/3}]$. The product $P(x)Q(x)$ will then satisfy all your conditions.

**Part 2: $n<cN^{1/3}$ is insufficient**

One thing that is written in every decent textbook addressing such inequalities is that, in principle, you need no fancy tools for them. Everything can be derived directly from the Lagrange interpolation formula **if** you can guess the right nodes to use. This is just the general linear programming mumbo-jumbo.

So, the question is how to get the right nodes in your setting. If you do not care about constant factors, then the canonical set of nodes for human consumption is just $\{k^2:k=0,1,\dots,n\}$. Let's recall how it works. Suppose you know the values $P(0),P(1),P(4),\dots,P(n^2)$ of a polynomial of degree $N$ and want to estimate $P'(-y), 0<y<1$ or something like that. You just write 
$$
P(x)=\sum{k=0^n}P(k)L_k(x)
$$
where
$$
L_k(x)=\frac {\prod_{m:\ne k}(x-m^2)}{\prod_{m:m\ne k}(k^2-m^2)}
$$
and estimate 
$$
|P'(-y)|\le\max_k|P(k^2)|\sum_k |L'_k(-y)|
$$
Now let us estimate $|L'_k(-y)|$. It is just
$$
|L_k(-y)|\sum_{m:m\ne k}\frac 1{y+m^2}
$$
Now, 
$$
|L_k(y)|=\frac{\prod_{m:m\ne k}(y+m^2)}{\prod_{m:m\ne k}(|m-k|(m+k))}
$$
If $k=0$, then the numerator is $\le (n!)^2\prod_{m=1}^n(1+\frac y{m^2})\le C(n!)^2$ while the denominator is exactly $(n!)^2$. Thus $|L_0(-y)|\le C$ whence 
$$
|L_0'(-y)|\le C\sum_{m=1}^n\frac 1{y+m^2}\le C\,.
$$
If $k>0$, then the numerator in $L_k(-y)$ is at most $Cy\frac 1{k^2}(n!)^2$ (the same count except now $y=|-y-0|$ is present and $y+k^2=|-y-k^2|$ is missing) while the denominator is
$$
\frac{k!(n-k)!(n+k)!}{2k(k-1)!}=\frac{(n-k)!(n+k)!}{2}
$$
whence
$$
|L_k'(-y)|\le \left(\frac 1y+\sum_{m\ge 1}\frac 1{y+m^2}\right)|L_k(-y)|\le
\frac C{k^2}\frac{(n!)^2}{(n-k)!(n+k)!}\le \frac C{k^2}e^{-k^2/n}\,.
$$
Thus we can easily derive from here that 
$$
|P'(-y)|\le C\sum_{k=0}^n |P(k^2)|\frac 1{k^2+1}e^{-k^2/n}\,.
$$

To be continued...