NZMO Round Two

Solution:
Note that if $n = j$ works then $n > j$ also works for a positive integer $j$ as we can just set $x_{i} = 0$ for $n \geq i > j$ and have $x_{1}, \ldots, x_{n}$ be the sequence that worked for $n$.

Consider $n = 4$. We take $x_{1}, x_{2} = \sqrt[4]{8}$ and $x_{3}, x_{4} = -\sqrt[4]{8}$ and all conditions are satisfied. Thus all $n \geq 4$ works.

Now if $n = 2$, we note $|x_{1}|,|x_{2}|< 2$ and thus $x_{1}^{4},x_{2}^{4}< 16$ so $x_{1}^{4} + x_{2}^{4}< 32$. Contradiction.

If $n = 3$, then wlog $x_{1},x_{2}\geq 0$ and $x_{3}\leq 0$ (as flipping the signs won't affect any of the conditions).
Now as $x_{1} + x_{2} = - x_{3}< 2$, we get that $(x_{1} + x_{2})^{4}< 16$. But if we expand this out we get $16 > (x_{1} + x_{2})^{4} = x_{1}^{4} + 4x_{1}^{3}x_{2} + 6x_{1}^{2}x_{2}^{2} + 4x_{1}x_{2}^{3} + x_{2}^{4}\geq x_{1}^{4} + x_{2}^{4}$ (as $x_{1},x_{2}\geq 0$).
Thus $x_{1}^{4} + x_{2}^{4}< 16$ but also note that $x_{3}^{4}< 16$, thus $x_{1}^{4} + x_{2}^{4} + x_{3}^{4}< 32$. Contradiction.

Thus, a sequence only exists for integers $n\geq 4$.