$\begin{aligned} \frac{(1+i)^{\mathrm{n}}}{(1-\mathrm{i})^{\mathrm{n}-2}} &=\left(\frac{1+\mathrm{i}}{1-\mathrm{i}}\right)^{\mathrm{n}}(1-\mathrm{i})^{2} \\ &=\left(1+\mathrm{i}^{2}-2 \mathrm{i}\right)\left(\frac{1+\mathrm{i}}{1-\mathrm{i}}\right)^{\mathrm{n}} \\ &=(1-1-2 \mathrm{i})\left(\frac{1+\mathrm{i}}{1-\mathrm{i}}\right)^{\mathrm{n}} \\ &=(-2 \mathrm{i})\left(\frac{1+\mathrm{i}}{1-\mathrm{i}}\right)^{\mathrm{n}} \end{aligned}$
$$
\begin{aligned}
&=(-2 i)\left\{\frac{(1+i)(1+i)}{(1-i)(1+i)}\right\}^{n} \\
&=(-2 i)\left\{\frac{2 i}{2}\right\}^{n}\quad\left(\because \mathrm{i}^{2}=-1\right) \\
&=(-2 i)(i)^{\mathrm{n}} \\
&=(-2) i^{n+1}
\end{aligned}
$$
Put $\mathrm{n}=1$, we get
$$
\begin{aligned}
&=(-2) i^{2}=(-2)(-1) \\
&=2 \Rightarrow \text { positive integer. }
\end{aligned}
$$
Hence, the least positive integer value of $n=1$ for which the given expression have positive value.