67th Czech and Slovak Mathematical Olympiad

Such numbers don't exist. For the sake of contradiction, assume that there exist positive integers $n$, $k$, $a$ such that $$\frac{n}{11^k-n}=a^2$$ which rewrites as $$n(a^2+1)=a^2\cdot 11^k.$$ From $\text{GCD}(a^2,a^2+1)=1$ we deduce $a^2+1\mid 11^k$ and hence $a^2+1=11^t$ for $1\le t\le k$. In particular, $a^2\equiv 10(\mathrm{mod}11)$. However, this is impossible as the squares of integers give remainders $0,1,4,9,5,3,3,5,9,4,1,\dots$ upon division by $11$.