Let $\displaystyle n$ be the the smallest natural number which satisfies the condition. Above all, note that $\displaystyle x^{4} \equiv 0,1\pmod{16}$. Since $\displaystyle 1998\equiv 14\pmod{16}$, we can say that $\displaystyle n\geq 14$. Assume that $\displaystyle n=14$. Then we must have $\displaystyle 1998=a\cdot 5^{4} +b\cdot 3^{4} +c\cdot 1^{4} ,\ a+b+c=14$ for some $\displaystyle ( a,b,c) \in \mathbb{N}^{3}$ So we have $\displaystyle 124=39\cdot a+5\cdot b$. It is not difficult to verify this is impossible. On the other hand, $\displaystyle 1998=2\cdot 5^{4} +10\cdot 3^{4} +1\cdot 2^{4} +2\cdot 1^{4}$ Therefore, the answer is $\displaystyle n=15$.