$\newcommand{\de}{\delta}
\newcommand{\ep}{\epsilon}$

Note that $p\ln p-p$ is decreasing in $p\in[0,1]$, so that $p\ln p-p\le q\ln q-q$ and hence $p\ln p-q\ln q\le p-q=|p-q|$ if $0\le q\le p\le1$. 

Next, take any real $c\ge1$. Note that $g(p):=p\ln p+cp$ (with $g(0):=0$) is convex in $p\in[0,1]$. So, assuming $0\le p\le q\le1$ and $q\ge e^{-c}$, we have 
$g(p)\le g(0)\vee g(q)=0\vee g(q)=g(q)$ and hence 
$p\ln p-q\ln q\le c(q-p)=c|p-q|$. 
Thus, 
\begin{equation}
	p\ln p-q\ln q\le c|p-q|
\end{equation}
whenever $0\le p,q\le1$ and $q\ge e^{-c}$. 

Also, $-q\ln q$ is increasing in $q\in[0,e^{-1}]$ and hence in $q\in[0,e^{-c}]$, so that $-q\ln q\le ce^{-c}$ for $q\in[0,e^{-c}]$. Also, $p\ln p\le0$ if $0\le p\le1$. 

Therefore, the difference between the entropies of $Q=(q_i)_{i=1}^N$ and $P=(p_i)_{i=1}^N$ is 
\begin{equation}
	\sum_1^N p_i\ln p_i-\sum_1^N q_i\ln q_i=S_1+S_2+S_3,
\end{equation}
where 
\begin{align*}	
S_1	&:=\sum_{i\colon q_i\ge e^{-c}} (p_i\ln p_i-q_i\ln q_i)\le \sum_{i\colon q_i\ge e^{-c}}c|p_i-q_i|
\le c\de \quad\text{if}\ \de\ge\|P-Q\|_1, \\
S_2	&:= \sum_{i\colon q_i< e^{-c}}p_i\ln p_i\le0, \\ 
S_3	&:= \sum_{i\colon q_i< e^{-c}}(-q_i\ln q_i)\le\sum_{i\colon q_i< e^{-c}}ce^{-c}\le Nce^{-c}.  
\end{align*}
So, taking now $c=\ln\frac N\de$ and assuming $N\ge e\de$, we see that 
\begin{equation}
	\sum_1^N p_i\ln p_i-\sum_1^N q_i\ln q_i\le 
	c\de+Nce^{-c}
	=2\de\ln\frac N\de.
\end{equation}
Taking here $\de=2\ep$ and noting that $N\ge1$, we conclude that 
\begin{equation}
	\sum_1^N p_i\ln p_i-\sum_1^N q_i\ln q_i\le 
4\ep\ln\frac{N}{2\ep}
\end{equation}
if $\ep\le1/(2e)$.