For $n=1$, we have that $(x - y)^{5} + (y -z)^{5} + (z - x)^{5}=-5(x-y)(y-z)(z-x)(x^2+y^2+z^2-xy-yz-zx)$, so the condition is satisfied. Now, assume $n \geq 2$. We have that $f_n(x,x,0) = 2x^{2n}-x^2$ divides $g_n(x,x,0) = x^{5n}+(-x)^{5n}$. If $n\equiv 0 \bmod{2}$, we have: $$2x^{2n}-x^2 \mid 2x^{5n}$$which is obviously false for $n \geq 2$. Hence $n\equiv 1 \bmod{2}$. We also have that $f_n(x,-1,0) = x^{2n}+x+1$ divides $g_n(x,-1,0) = (x+1)^{5n}-x^{5n}-1$: $$x^{2n}+x+1 \mid (x+1)^{5n}-x^{5n}-1 \Longleftrightarrow$$$$x^{2n}+x+1 \mid \sum_{i=1}^{5n-1} \binom {5n}i \cdot x^i \Longleftrightarrow$$$$x^{2n}+x+1 \mid \sum_{i=1}^{2n-1} \binom {5n}i \cdot x^i- \sum_{i=2n}^{4n-1} \binom {5n}i \cdot x^{i-2n} \cdot (x+1)+\sum_{i=4n}^{5n-1} \binom {5n}i \cdot x^{i-4n} \cdot (x+1)^2 \Longleftrightarrow$$The degree of the $LHS$ is equal to $2n$ and so is the degree of the $RHS$. Hence, in order to obtain $RHS$, we must multiply the $LHS$ by a constant polynomial (i.e. an integer). This means that the coefficient of $x^{2n-1}$ in the $RHS$ is zero (since $n \geq 3$, $n$ is odd). The coefficient of $x^{2n-1}$ equals: $$\binom {5n}{2n-1}-\binom {5n}{4n-2}-\binom {5n}{4n-1}=\binom {5n}{2n-1}-\binom {5n}{n+2}-\binom {5n}{n+1}$$For $n=3$, we have that $\binom {5n}{2n-1}-\binom {5n}{n+2}-\binom {5n}{n+1}=\binom {15}{5}-\binom {15}{5}-\binom {15}{4}=-\binom {15}{4} \neq 0$, so $n >3$. I will prove that $$\binom {5n}{2n-1} > 2\binom {5n}{n+2}$$for every $n \geq5$, so we will derive a contradiction with: $$\binom {5n}{2n-1} > 2\binom {5n}{n+2} > \binom {5n}{n+2}+\binom {5n}{n+1}$$We proceed with induction on $n$: For $n=5$, we can manually check the inequality. For the inductive step, notice that the inductive hypothesis is equivalent to $$(4n-2)!(n+2)!>2(2n-1)!(3n+1)!$$and the desired result is $$(4n+2)!(n+3)!>2(2n+1)!(3n+4)! \Longleftrightarrow$$$$(4n-2)!(4n-1)(4n)(4n+1)(4n+2)(n+2)!(n+3)>2(2n-1)!(2n)(2n+1)(3n+1)!(3n+2)(3n+3)(3n+4)$$Because of the inductive hypothesis, it suffices to show that: $$(4n-1)(4n)(4n+1)(4n+2)(n+3)>(2n)(2n+1)(3n+2)(3n+3)(3n+4) \Longleftrightarrow$$$$4(4n-1)(4n+1)(n+3)>(3n+2)(3n+3)(3n+4) \Longleftrightarrow$$Because $(3n+2)(3n+3)(3n+4)<(3n+2)(3n+3)(3n+9)$, it suffices to show: $$4(4n-1)(4n+1)>3(3n+2)(3n+3) \Longleftrightarrow$$$$64n^2-4>27n^2+45n+18 \Longleftrightarrow$$$$37n^2-45n-22>0 \Longleftrightarrow$$$$n(37n-45)>22$$which is true for $n \geq 5$. We have reached a contradiction for $n \geq 2$, so the only positive integer satisfying the conditions of the problem is $n=1$