Any solution?

trivial problem my solution: obviously $1,2,3,4,5$ aren't crazy so among $1,2,\cdots ,2014$ there are less than $2012$ crazy numbers.now let $n=a^{b(b+1)(b+2)\cdots (b+2013)}$ note that $n+b,n+b+1,n+b+2,\cdots n+b+2013$ are $2014$ consecutive crazy numbers so using continuity principle we get that there are $2014$ consecutive natural numbers such that among them are exactly $2012$ crazy numbers.

I don't quite follow, how did you show there are *exactly* 2012 crazy numbers?

Let the number of crazy numbers between $n, n+1, \cdots n+2013$ be $f(n)$. Now since $1,2,3,4,5$ are not crazy, we have $f(1) < 2012$. Also, using andria's construction, we have that there exists an $n$ such that $f(n)=2014$. Now note that $f(x+1)-f(x) \in \{-1,0,1\}$, so to go from $f(1)<2012$ to $f(n)=2014$, there must be an $i \in (1,n)$ such that $f(i)=2012$, as desired.

Ah okay thanks!

Solved with Alan Bu, Alex Zhao, Christopher Qiu, Edward Yu, Eric Shen, Isaac Zhu, Jeffrey Chen, Kevin Wu, and Ryan Yang. Consider the set $\{2^{2014!}+1, 2^{2014!}+2, \cdots, 2^{2014!}+2014\}$, which contains $2014$ naughty numbers \[2^{1\cdot 2014!}+1, 2^{2 \cdot \frac{2014!}{2}}+2,\cdots. \]Let $f(n)$ be the number of crazy numbers from $n$ to $n+2013$. Notice that $|f(n)-f(n-1)|\leq 1$. It can easily be checked that there are fewer than $2012$ crazy numbers from $1$ to $2014$, so by "discrete continuity" of $f$ we have that there exists some $n<2^{2014!}+1$ which satisfies the problem condition.

$A=a^{n!}+k$ for $k=2,3,...,n$ $A$ is crazy number.

$2^{2015!}+k$ $k=2,3 \dots 2015$ contains 2014 crazy numbers, so now we keep decreasing or "shifting" these by $1$, each time the no. of crazy no. changes by $\leq1$ and at the end there when we reach at ${1,2,..2014}$ there are less than $2014$ crazy no. , hence, by "discrete continuity/nullasz" we are done.