Showing an identity between polynomials whose coefficients involve combinatorial identities I want to show that
$$
 \sum_{k=0}^{\lfloor n/2 \rfloor} \sum_{l=0}^{\lfloor \frac{n-2k}{2}\rfloor} (-1)^l \binom{n}{k} \binom{n-2k-l}{l} \frac{n-2k}{n-2k-l} x^{n-2k-2l} = x^n.
$$
If we "compare coefficients", then we get $x^n$ on the left if $k = l = 0$, hence 
$$
 \binom{n}{0} \binom{n}{0} \frac{n}{n} x^{n} = x^n
$$
is valid. But the other sums look quite messy...
 A: Putting $n=2N$ we want to show for integral $N\geq 0$:
\begin{align*}
\sum_{k=0}^N\sum_{l=0}^{N-k}(-1)^l\binom{2N}{k}\binom{2N-2k-l}{l}
\frac{2N-2k}{2N-2k-l}x^{2N-2k-2l}=x^{2N}
\end{align*}

At first we derive a somewhat more convenient representation.
  We obtain
  \begin{align*}
\color{blue}{\sum_{k=0}^N}&\color{blue}{\sum_{l=0}^{N-k}(-1)^l\binom{2N}{k}\binom{2N-2k-l}{l}
\frac{2N-2k}{2N-2k-l}x^{2N-2k-2l}}\\
&=\sum_{k=0}^N\sum_{l=0}^{N-k}(-1)^{N-k-l}\binom{2N}{k}\binom{N-k+l}{N-k-l}
\frac{2N-2k}{N-k+l}x^{2l}\tag{1}\\
&=\sum_{k=0}^N\sum_{l=0}^{k}(-1)^{k-l}\binom{2N}{N-k}\binom{k+l}{k-l}
\frac{2k}{k+l}x^{2l}\tag{2}\\
&=\sum_{l=0}^N\sum_{k=l}^{N}(-1)^{k-l}\binom{2N}{N-k}\binom{k+l}{k-l}
\frac{2k}{k+l}x^{2l}\tag{3}\\
&\,\,\color{blue}{=\sum_{l=0}^N\sum_{k=0}^{N-l}(-1)^{k}\binom{2N}{N-k-l}\binom{k+2l}{k}
\frac{2k+2l}{k+2l}x^{2l}}\tag{4}\\
\end{align*}

Comment: 


*

*In (1) we exchange the summation order of the inner sum $l\to N-k-l$.

*In (2) we exchange the summation order of the outer sum $k\to N-k$.

*In (3) we exchange the series.

*In (4) we shift the index of the inner sum to start with $k=0$.

Let's call the polynomial $P(x)$. We can easily derive from (4) the  coefficient $[x^{2N}]$ of the highest power and see
  \begin{align*}
\color{blue}{[x^{2N}]P(x)=1}
\end{align*}
Next we consider for integral $N>0$ and $0\leq t<N$ according to (4):
  \begin{align*}
&\color{blue}{[x^{2t}]P(x)=\sum_{k=0}^{N-t}(-1)^{k}\binom{2N}{N-k-t}\binom{k+2t}{k}
\frac{2k+2t}{k+2t}}\\
&\quad=\sum_{k=0}^{N-t}(-1)^{k}\binom{2N}{N-k-t}\binom{k+2t}{k}
\left(1+\frac{k}{k+2t}\right)\\
&\quad=\sum_{k=0}^{N-t}(-1)^{k}\binom{2N}{N-k-t}\binom{k+2t}{k}+\sum_{k=1}^{N-t}(-1)^{k}\binom{2N}{N-k-t}\binom{k+2t-1}{k-1}\tag{5}\\
&\quad=\sum_{k=0}^{N-t}(-1)^{k}\binom{2N}{N-k-t}\binom{k+2t}{k}
+\sum_{k=0}^{N-t-1}(-1)^{k}\binom{2N}{N-k-t-1}\binom{k+2t}{k}\tag{6}\\
&\quad=\sum_{k=0}^{N-t}\binom{2N}{N-k-t}\binom{-2t-1}{k}
+\sum_{k=0}^{N-t-1}\binom{2N}{N-k-t-1}\binom{-2t-1}{k}\tag{7}\\
&\quad=\binom{2N-2t-1}{N-t}-\binom{2N-2t-1}{N-t-1}\tag{8}\\
&\quad\,\,\color{blue}{=0}
\end{align*}
  and the claim follows.

Comment:


*

*In (5) we multiply out and apply to the right-hand series the binomial identity $\binom{p}{q}=\frac{p}{q}\binom{p-1}{q-1}$.

*In (6) we shift the index of the right-hand series to start with $k=0$.

*In (7) we apply the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$ twice.

*In (8) we apply the Chu-Vandermonde Identity twice.
