We have $z=-x-y$. Denote $d=\gcd(x,y)$, $x=da$, $y=db$, so $\gcd(a,b)=1$. Then $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{n} $ writes as $n = \frac {dab(a+b)} {a^2+ab+b^2}$. But $\gcd(a,a^2+ab+b^2) = \gcd(b,a^2+ab+b^2) = \gcd(a+b,a^2+ab+b^2) = 1$, so we need $a^2+ab+b^2 \mid d$, when we get $n = \frac {d} {a^2+ab+b^2} ab(a+b)$. But since $ab(a+b)$ is always even, $n$ will be even. Conversely, for $n=2k$, let us take $a=b=1$, $d=3k$, so $x=y=3k$, $z=-6k$, for which $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{3k}+\frac{1}{3k}-\frac{1}{6k}=\frac{1}{2k}=\frac{1}{n} $, hence the answer is all $n$ even.

I got a slightly different solution. First, consider mavropnevma's example to prove that $n$ even is possible. For the other part, let $2^k$ be the greatest power of $2$ that divides all of $x,y,z$, and let $x/2^k = x'$, $y/2^k = y'$ and $z/2^k = z'$. Since $x'+ y'+ z'= 0$ and not all of them are even, exactly one of $x',y',z'$ is even. Suppose wlog $x'$ is even, that is, $x' = 2t$. Then \[\frac 1x + \frac 1y + \frac 1z = \frac1{2^k}\left(\frac 1{2t} + \frac1{y'} + \frac1{z'}\right) = \frac{y'z'+2t(y'+z')}{2^{k+1}ty'z'}\] and since $y'z' + 2t(y'+z')$ is odd, one cannot simplify $2^{k+1}$ in the denominator, so the denominator of $\frac1x + \frac1y + \frac 1z$ is always even.

For storage. We are given with constraint $x+y+z=0$:$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{n}$$Set $z=-x-y$ and obtain$$\frac{y(-x-y)+x(-x-y)+xy}{xy(-x-y)}=\frac{1}{n}\Longleftrightarrow n=\frac{xy(x+y)}{x^2+xy+y^2}.$$If $x=y$, then $n=\frac{2x}{3}$, thus we can achieve every $n$ that is even by setting $x=y=3k$ and $z=-x-y=-6k$, where $k\in\mathbb Z$. Now let us show that odd $n$ is not achievable. Note that if $x$ is odd, $y$ even or if $x$ is even, $y$ is odd, then numerator is even and denominator is odd, thus if $n$ is integer, then $2\mid n$. If both $x,y$ are odd, then numerator is even, since $x+y$ is even and denominator is odd, thus if $n$ is integer, then $2\mid n$. Now suppose both $x,y$ are even, then let $x=2m$ and $y=2n$ and we achieve$$n=\frac{2mn(m+n)}{m^2+mn+n^2},$$we have reduced our condition, but now it is a multiple of $2$, repeat same arguments and therefore we have shown that only even $n$ is possible. Answer. $\boxed{n=2k\forall k\in \mathbb Z\neq 0}$.

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