XIV OBM

We only need to prove that there are positive integers $n$, $k$ such that $$\begin{gathered} \underset{1992}{\underbrace{11\dots 11}}\cdot 10^k\le n^{1992}<\underset{1991}{\underbrace{11\dots 1}}2\cdot 10^k \\ ⟺\sqrt[1992]{11\dots 11}\cdot 10^t\le n<\sqrt[1992]{11\dots 12}\cdot 10^t \end{gathered}$$ in which $k=1992t$. Suppose the decimal expansions of $\sqrt[1992]{11\dots 11}$ and $\sqrt[1992]{11\dots 12}$ differ in the $t$th position. So $$n=\lceil \sqrt[1992]{\underset{1992}{\underbrace{11\dots 11}}}\cdot 10^t\rceil +1$$ suffices.