14. 5. 11 $\star \star$ Find the positive integer $n$ that satisfies $133^{5}+110^{5}+84^{5}+27^{5}=n^{5}$.
Because $n^{5}=133^{5}+110^{5}+84^{5}+27^{5} \equiv 4(\bmod 10)$, so the unit digit of $n$ is 4, and $n^{5} \equiv 0(\bmod 3)$, so $3 \mid n$. The numbers that meet these conditions and are greater than 133 are $144,174, \cdots$, and $174^{5}=133^{5} \times\left(\frac{174}{133}\right)^{5}>133^{5} \times 1.3^{5}>133^{5} \times 3>133^{5}+110^{5}+(84+27)^{5}$, so $n=144$.