Let $w,x,y,$ and $z$ be nonzero complex numbers, and let $n$ be a positive integer. Suppose that the following conditions hold:

[list]
    [*] $\frac{1}{w}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3,$

    [*] $wx+wy+wz+xy+xz+yz=14,$
    
    [*] $(w+x)^3+(w+y)^3+(w+z)^3+(x+y)^3+(x+z)^3+(y+z)^3=2160, \text{ and}$
    
    [*] $w+x+y+z+i\sqrt{n} \in \mathbb{R}.$
[/list]

Compute $n$.

[i]Proposed by Luke Robitaille[/i]
Given the conditions:
1. \(\frac{1}{w} + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 3\),
2. \(wx + wy + wz + xy + xz + yz = 14\),
3. \((w+x)^3 + (w+y)^3 + (w+z)^3 + (x+y)^3 + (x+z)^3 + (y+z)^3 = 2160\),
4. \(w + x + y + z + i\sqrt{n} \in \mathbb{R}\).

Let \(s = w + x + y + z\). We need to use the given conditions to find \(n\).

First, consider the third condition. We can use the identity for the sum of cubes:
\[
(a+b)^3 = a^3 + b^3 + 3ab(a+b).
\]
Applying this to each pair, we get:
\[
(w+x)^3 + (w+y)^3 + (w+z)^3 + (x+y)^3 + (x+z)^3 + (y+z)^3.
\]

We can rewrite the sum of cubes as:
\[
(w+x)^3 + (w+y)^3 + (w+z)^3 + (x+y)^3 + (x+z)^3 + (y+z)^3 = 2160.
\]

Using the identity for the sum of cubes, we have:
\[
(w+x)^3 = w^3 + x^3 + 3wx(w+x),
\]
\[
(w+y)^3 = w^3 + y^3 + 3wy(w+y),
\]
\[
(w+z)^3 = w^3 + z^3 + 3wz(w+z),
\]
\[
(x+y)^3 = x^3 + y^3 + 3xy(x+y),
\]
\[
(x+z)^3 = x^3 + z^3 + 3xz(x+z),
\]
\[
(y+z)^3 = y^3 + z^3 + 3yz(y+z).
\]

Summing these, we get:
\[
(w+x)^3 + (w+y)^3 + (w+z)^3 + (x+y)^3 + (x+z)^3 + (y+z)^3 = 2(w^3 + x^3 + y^3 + z^3) + 3(wx(w+x) + wy(w+y) + wz(w+z) + xy(x+y) + xz(x+z) + yz(y+z)).
\]

Given that \(wx + wy + wz + xy + xz + yz = 14\), we can simplify the expression:
\[
2160 = 2(w^3 + x^3 + y^3 + z^3) + 3 \cdot 14 \cdot s.
\]

Next, we use the fact that \(w + x + y + z + i\sqrt{n} \in \mathbb{R}\). This implies that \(s + i\sqrt{n} \in \mathbb{R}\), so \(i\sqrt{n}\) must be the imaginary part of \(s\). Therefore, \(s\) must be purely imaginary, and \(s = -i\sqrt{n}\).

Substituting \(s = -i\sqrt{n}\) into the equation:
\[
2160 = 3s^3 - 3s \cdot 28.
\]

Simplifying, we get:
\[
2160 = 3(-i\sqrt{n})^3 - 3(-i\sqrt{n}) \cdot 28,
\]
\[
2160 = 3(-i^3 n^{3/2}) - 3(-i\sqrt{n}) \cdot 28,
\]
\[
2160 = 3(i n^{3/2}) + 84i\sqrt{n},
\]
\[
2160 = 3i n^{3/2} + 84i\sqrt{n}.
\]

Since \(2160\) is a real number, the imaginary part must be zero:
\[
3i n^{3/2} + 84i\sqrt{n} = 0,
\]
\[
i(3n^{3/2} + 84\sqrt{n}) = 0,
\]
\[
3n^{3/2} + 84\sqrt{n} = 0,
\]
\[
3n^{3/2} = -84\sqrt{n},
\]
\[
n^{3/2} = -28\sqrt{n}.
\]

Dividing both sides by \(\sqrt{n}\):
\[
n = -28.
\]

Since \(n\) must be a positive integer, we conclude that there is no such \(n\) that satisfies all the given conditions.

The final answer is \(\boxed{None}\).