Problem 1028

We will first give an example of 2012 consecutive quirky numbers. It suffices to take the numbers $N+2, N+3, \ldots, N+2013$, where $N=2^{2013!}$.
For a natural number $n$, let us denote by $f(n)$ the number of quirky numbers among $n, n+1, \ldots, n+2013$. Since $f(1)<2012$ (the numbers $1,2,3,4,5$ are not quirky), $f(N) \geqslant 2012$ and $|f(n+1)-f(n)| \leqslant 1$ for every $n$, there exists $n$ such that $f(n)=2012$.

Second solution. Let us denote $N=\frac{2014!}{2011}$. The numbers $2^{N}+i$ for $2 \leqslant i \leqslant 2010$ and $2012 \leqslant i \leqslant 2014$ are quirky; let us prove that $2^{N}+1$ and $2^{N}+2011$ are not.
Suppose that $2^{N}+2011=a^{b}+b$ (the number $2^{N}+1$ is examined similarly). From $2^{b} \leqslant a^{b}<2^{N}$ it follows that $b<N$. Also $b>2011$, since otherwise we would have $2011-b=a^{b}-2^{N}=\left(a-2^{\frac{N}{b}}\right)\left(\cdots+2^{\frac{(b-1) N}{b}}\right)>2^{\frac{N}{b}}>2011$. Further, if $2 \mid a$, we have $2^{b} \mid 2^{N}-a^{b}=b-2011<2^{b}$, which is impossible. Finally, for $2 \nmid a$ and $2 \mid b$ we have $b-2011=\left(2^{\frac{N}{2}}-a^{\frac{b}{2}}\right)\left(2^{\frac{N}{2}}+a^{\frac{b}{2}}\right)>2^{\frac{N}{2}}$, so $2^{N}+2011=a^{b}+b>2^{2^{N / 2}}$, again impossible since $2^{\frac{N}{2}}>N$.