已知在 $\left(\sqrt{x} - \frac{2}{x}\right)^n$ 的展开式中存在关于 $x^2$ 的项，其中 $n$ 为正整数。

1. Write down the general term of the binomial expansion

The general term is given by:
$T_{r+1} = \binom{n}{r} (\sqrt{x})^{n-r} \left(-\frac{2}{x}\right)^r$

2. Simplify the general term to find the power of $x$

$T_{r+1} = \binom{n}{r} x^{\frac{n-r}{2}} (-2)^r x^{-r}$
$T_{r+1} = \binom{n}{r} (-2)^r x^{\frac{n-r}{2} - r}$
$T_{r+1} = \binom{n}{r} (-2)^r x^{\frac{n-3r}{2}}$

3. Equate the power of $x$ to 2 to find the relationship between $n$ and $r$

For a term in $x^2$, the power of $x$ must be 2:
$\frac{n-3r}{2} = 2$
$n - 3r = 4$
$n = 3r + 4$

4. Determine the smallest possible value of $n$ with explanation

Since $r$ must be a non-negative integer ($r \geqslant 0$), the smallest possible value of $n$ occurs when $r = 0$.
$n = 3(0) + 4 = 4$