Existence / non-existence of a sequence with short longest increasing subsequence and decreasing subsequence?
The answer to the OP's question is, no if $N\le 7$ and yes otherwise.

For given any positive integer $r$ and $s$, the celebrated Erdős–Szekeres theorem shows that for any sequence of distinct real numbers with length at least $(r − 1)(s − 1) + 1$ contains an increasing subsequence of length $r$ or a decreasing subsequence of length $s$.
It turns out that bound, $(r-1)(s-1)+1$ is tight. That is, for any positive number $r$ and $s$, there is a sequence of distinct numbers with length $(r-1)(s-1)$ that contains no increasing subsequence of length $r$ and no decreasing subsequence of length $s$.
Here is such an example.
$$\begin{array} {}
&amp;s-1, &amp;s-2, &amp;\cdots,&amp;2, &amp;1\\
&amp;2(s-1), &amp;(s-1)+ s-2,  &amp;\cdots, &amp;(s-1)+ 2, &amp;(s-1)+ 1\\
&amp;\vdots &amp;\vdots &amp;\vdots &amp;\vdots &amp;\vdots \\
&amp;(r-2)(s-1), &amp;(r-3)(s-1)+s-2, &amp;\cdots, &amp;(r-3)(s-1)+2, &amp;(r-3)(s-1)+1\\
&amp;(r-1)(s-1), &amp;(r-2)(s-1)+s-2, &amp;\cdots, &amp;(r-2)(s-1)+2, &amp;(r-2)(s-1)+1\\
\end{array}$$
Consider the numbers above, reading from left to right and then from top to bottom. In other words, the sequence is $s-1$ down to $1$, followed by $2(s-1)$ down to $(s-1)+1$, etc and finally followed by $(r-1)(s-1)$ down to $(r-2)(s-1)+1$, all in step of $1$.
It is easy to see that there is no increasing subsequence of length r and no decreasing subsequence of length $s$.
For example, when $r=s=5$, we have $$4,3,2,1,\ \, 8,7,6,5,\ \,12,11,10,9,\ \,16,15,14,13$$
which does not have increasing subsequence of length $5$ nor decreasing subsequence of length $5$.

If we let $r=s$, the section above implies that, for any positive number $N$, there exists an integer sequence of length $N$ with all unique elements such that the length of its longest increasing subsequence as well as that of its longest decreasing subsequence is at most $\lceil\sqrt N\rceil$. And $\lceil\sqrt N\rceil$ is the tight upper bound.
Since $$\lceil\sqrt N\rceil\ge \lfloor\frac N2\rfloor\ \text{ for all } N\le 7$$ and $$\lceil\sqrt N\rceil\lt \lfloor\frac N2\rfloor\ \text{ for all } N\gt 7,$$
the answer to the OP's question is, no if $N\le 7$ and yes otherwise.
For example, for $N=8$, we have sequence $3,2,1,6,5,4,9,8,7$.