The Guardians!

We will show that if $n$ is an odd positive integer then

$\large\ \sum _{ k=1 }^{ n-1 }{ { \left( \tan { \left( \frac { k\pi }{ n } \right) } \right) }^{ 4 } } = 2{n \choose 2}^2 - 4{n \choose 4}.$

and for $n =2019$ , it gives $553890090097$ which has indeed $\boxed{4}$ zeroes.

It is easy to verify that

$\large\ \displaystyle { \prod _{ k=0 }^{ n-1 }{ \left( x - { t }_{ k } \right) } = Re\left( { \left( x+i \right) }^{ n } \right) = \sum _{ k=0 }^{ \frac { \left( n - 1 \right) }{ 2 } }{{n \choose {2k} } { x }^{ n-2k }{ \left( -1 \right) }^{ k } } },$

where ${ t }_{ k } = \tan { \left( \frac { k\pi }{ n } \right) }$ , since both sides are monic polynomials of degree $n$ which vanish at $x = t_k.$ Hence, letting

$\large\ {\displaystyle \prod _{ k=0 }^{ n-1 }{ \left( { x }^{ 4 }-{ { { t }_{ k } }^{ 4 } } \right) } = -\prod _{ k=0 }^{ n-1 }{ \left( x-{ t }_{ k } \right) \left( -x-{ t }_{ k } \right) \left( ix-{ t }_{ k } \right) \left( -ix-{ t }_{ k } \right) } ={ x }^{ 4n } - { \left( 2{n \choose 2}^2 - 4{n \choose 4} \right)}{ x }^{ 4n-4 } + R(x) },$

where $R(x)$ is the sum of all the terms of lower degree. On the other hand

$\large\ { \prod _{ k=0 }^{ n-1 }{ \left( { x }^{ 4 } - { { { t }_{ k } }^{ 4 } } \right) } = x^{4n} -{S_n}x^{4n-4} + R(x) }$

and the desired result follows.