Here is the proof of Kevin Liu's version $$ \sum_{k=1}^{2n+1}\left(\frac{(-y)^k}{1-y^{3k}}+ \frac{y^{k}}{1+y^{3k}}\right)=-n-1 $$ (for the primitive root of unity $y$ of degree $6n+4$) of Nemo's reduction. (Both reductions deserve to be explained, in my opinion.)

We have $$\sum_{k=1}^{2n+1} \frac{(-y)^k}{1-y^{3k}}= -\sum_{k=1}^{2n+1} \frac{y^{k}}{1-y^{3k}}+2\sum_{k=1}^n\frac{y^{2k}}{1-y^{6k}}.$$

So we should prove

$$2\sum_{k=1}^{2n+1} \frac{y^{4k}}{1-y^{6k}}-2\sum_{k=1}^n \frac{y^{2k}}{1-y^{6k}}=n+1$$

Denote again $w=y^2$, the primitive root of unity of degree $3n+2$, this reads as

$$ 2\sum_{k=1}^{2n+1}\frac{w^{2k}}{1-w^{3k}}-2\sum_{k=1}^n\frac{w^k}{1-w^{3k}}=n+1 $$

Partition LHS onto two identical halfs (it has multiple 2 for that), and in one of them make the change of variables $k\mapsto 3n+2-k$. This half reads as $$ \sum_{k=n+1}^{3n+2}\frac{w^{-2k}}{1-w^{-3k}}-\sum_{k={2n+2}}^{3n+1}\frac{w^{-k}}{1-w^{-3k}}= -\sum_{k=n+1}^{3n+2}\frac{w^{k}}{1-w^{3k}}+\sum_{k={2n+2}}^{3n+1}\frac{w^{2k}}{1-w^{3k}}. $$ Collecting with another half we get (so lucky) $$ \sum_{k=1}^{3n+1} \frac{w^{2k}-w^k}{1-w^{3k}}=-\sum_{k=1}^{3n+1} w^k\frac{1-w^{3k(n+1)}}{1-w^{3k}}= -\sum_{k=1}^{3n+1} (w^k+w^{4k}+w^{7k}+\dots+w^{(3n+1)k})=n+1, $$ since the numbers $1,4,\dots,3n+1$ are not divisible by $3n+2$ and $\sum_{k=0}^{3n+1} w^{kd}=0$ for all integer $d$ non-divisible by $3n+2$.

Here is the proof of Kevin Liu's version $$ \sum_{k=1}^{2n+1}\left(\frac{(-y)^k}{1-y^{3k}}+ \frac{y^{k}}{1+y^{3k}}\right)=-n-1 $$ (for the primitive root of unity $y$ of degree $6n+4$) of Nemo's reduction. (Both reductions deserve to be explained, in my opinion.)

We have $$\sum_{k=1}^{2n+1} \frac{(-y)^k}{1-y^{3k}}= -\sum_{k=1}^{2n+1} \frac{y^{k}}{1-y^{3k}}+2\sum_{k=1}^n\frac{y^{2k}}{1-y^{6k}}.$$

So we should prove

$$2\sum_{k=1}^{2n+1} \frac{y^{4k}}{1-y^{6k}}-2\sum_{k=1}^n \frac{y^{2k}}{1-y^{6k}}=n+1$$

Denote again $w=y^2$, the primitive root of unity of degree $3n+2$, this reads as

$$ 2\sum_{k=1}^{2n+1}\frac{w^{2k}}{1-w^{3k}}-2\sum_{k=1}^n\frac{w^k}{1-w^{3k}}=n+1 $$

Partition LHS onto two identical halfs (it has multiple 2 for that), and in one of them make the change of variables $k\mapsto 3n+2-k$. This half reads as $$ \sum_{k=n+1}^{3n+1}\frac{w^{-2k}}{1-w^{-3k}}-\sum_{k={2n+2}}^{3n+1}\frac{w^{-k}}{1-w^{-3k}}= -\sum_{k=n+1}^{3n+1}\frac{w^{k}}{1-w^{3k}}+\sum_{k={2n+2}}^{3n+1}\frac{w^{2k}}{1-w^{3k}}. $$ Collecting with another half we get (so lucky) $$ \sum_{k=1}^{3n+1} \frac{w^{2k}-w^k}{1-w^{3k}}=-\sum_{k=1}^{3n+1} w^k\frac{1-w^{3k(n+1)}}{1-w^{3k}}= -\sum_{k=1}^{3n+1} (w^k+w^{4k}+w^{7k}+\dots+w^{(3n+1)k})=n+1, $$ since the numbers $1,4,\dots,3n+1$ are not divisible by $3n+2$ and $\sum_{k=0}^{3n+1} w^{kd}=0$ for all integers $d$ non-divisible by $3n+2$.

Here is the proof of Kevin Liu's version $$ \sum_{k=1}^{2n+1}\left(\frac{(-y)^k}{1-y^{3k}}+ \frac{y^{k}}{1+y^{3k}}\right)=-n-1 $$ of Nemo's reduction (both reductions deserve to be explained, in my opinion.)

We have $$\sum_{k=1}^{2n+1} \frac{(-y)^k}{1-y^{3k}}= -\sum_{k=1}^{2n+1} \frac{y^{k}}{1-y^{3k}}+2\sum_{k=1}^n\frac{y^{2k}}{1-y^{6k}}.$$

So we should prove

$$2\sum_{k=1}^{2n+1} \frac{y^{4k}}{1-y^{6k}}-2\sum_{k=1}^n \frac{y^{2k}}{1-y^{6k}}=n+1$$

Denote again $w=y^2$, the primitive root of unity of degree $3n+2$, this reads as

$$ 2\sum_{k=1}^{2n+1}\frac{w^{2k}}{1-w^{3k}}-2\sum_{k=1}^n\frac{w^k}{1-w^{3k}}=n+1 $$

Partition LHS onto two identical halfs (it has multiple 2 for that), and in one of them make the change of variables $k\mapsto 3n+2-k$. This half reads as $$ \sum_{k=n+1}^{3n+2}\frac{w^{-2k}}{1-w^{-3k}}-\sum_{k={2n+2}}^{3n+1}\frac{w^{-k}}{1-w^{-3k}}= -\sum_{k=n+1}^{3n+2}\frac{w^{k}}{1-w^{3k}}+\sum_{k={2n+2}}^{3n+1}\frac{w^{2k}}{1-w^{3k}}. $$ Collecting with another half we get (so lucky) $$ \sum_{k=1}^{3n+1} \frac{w^{2k}-w^k}{1-w^{3k}}=-\sum_{k=1}^{3n+1} w^k\frac{1-w^{3k(n+1)}}{1-w^{3k}}= -\sum_{k=1}^{3n+1} (w^k+w^{4k}+w^{7k}+\dots+w^{(3n+1)k})=n+1, $$ since the numbers $1,4,\dots,3n+1$ are not divisible by $3n+2$ and $\sum_{k=0}^{3n+1} w^{kd}=0$ for all integer $d$ non-divisible by $3n+2$.

Here is the proof of Kevin Liu's version $$ \sum_{k=1}^{2n+1}\left(\frac{(-y)^k}{1-y^{3k}}+ \frac{y^{k}}{1+y^{3k}}\right)=-n-1 $$ (for the primitive root of unity $y$ of degree $6n+4$) of Nemo's reduction. (Both reductions deserve to be explained, in my opinion.)

We have $$\sum_{k=1}^{2n+1} \frac{(-y)^k}{1-y^{3k}}= -\sum_{k=1}^{2n+1} \frac{y^{k}}{1-y^{3k}}+2\sum_{k=1}^n\frac{y^{2k}}{1-y^{6k}}.$$

So we should prove

$$2\sum_{k=1}^{2n+1} \frac{y^{4k}}{1-y^{6k}}-2\sum_{k=1}^n \frac{y^{2k}}{1-y^{6k}}=n+1$$

Denote again $w=y^2$, the primitive root of unity of degree $3n+2$, this reads as

$$ 2\sum_{k=1}^{2n+1}\frac{w^{2k}}{1-w^{3k}}-2\sum_{k=1}^n\frac{w^k}{1-w^{3k}}=n+1 $$

Partition LHS onto two identical halfs (it has multiple 2 for that), and in one of them make the change of variables $k\mapsto 3n+2-k$. This half reads as $$ \sum_{k=n+1}^{3n+2}\frac{w^{-2k}}{1-w^{-3k}}-\sum_{k={2n+2}}^{3n+1}\frac{w^{-k}}{1-w^{-3k}}= -\sum_{k=n+1}^{3n+2}\frac{w^{k}}{1-w^{3k}}+\sum_{k={2n+2}}^{3n+1}\frac{w^{2k}}{1-w^{3k}}. $$ Collecting with another half we get (so lucky) $$ \sum_{k=1}^{3n+1} \frac{w^{2k}-w^k}{1-w^{3k}}=-\sum_{k=1}^{3n+1} w^k\frac{1-w^{3k(n+1)}}{1-w^{3k}}= -\sum_{k=1}^{3n+1} (w^k+w^{4k}+w^{7k}+\dots+w^{(3n+1)k})=n+1, $$ since the numbers $1,4,\dots,3n+1$ are not divisible by $3n+2$ and $\sum_{k=0}^{3n+1} w^{kd}=0$ for all integer $d$ non-divisible by $3n+2$.