Combinatorics problem IMO style, existence type problem I've tried and failed solving this problem for a week now; it doesn't seem too difficult and I feel like I'm missing something obvious.  What is the solution?

EDIT:
I tried constructing some sequences that would work and would divide 2001, for example 23 divides 2001 and if you can make a sequence of coins with length 21 you could just use the sequence again and again, but I couldn't find such sequence that wouldn't interfere itself (by for example having a 3 on the outside); also I just noticed that my calculations were wrong and the length of the sequence doesn't perfectly divide 2001. I tried induction and strong induction but that didn't work; finally I tried using the hostile neighbors principle (which I read about in a book and I'm pretty sure the author made the name up)
 A: Hints:

*

*Notice that any $4$ consecutive coins can have at most one $3$-kopeyka.


*Show that if you choose any $4$ consecutive coins, say, $a_k, a_{k+1}, a_{k+2}, a_{k+3}$, then exactly one of them is $3$-kopeyka.
A: On the one hand, observe that there can be no arrangement of 4 or more consecutive coins that do not have a value of 3. Equivalently, let $a_j$ be the value of the $j$-th coin in the sequence; $j=1,2,\ldots,2001$. Then for each $i=0,1,\ldots,1997$ at least one of $a_{i+1},a_{i+2},a_{i+3},a_{i+4}$  must be 3. So the number $X$ of $j \in \{1,2,\ldots, 2001\}$ such that $a_j =$ 3 satisfies the inequality $X \ \ge \lfloor \frac{2001}{4} \rfloor$.
On the other hand, the condition that at least 3 other coins are between any coin of value 3 gives the inequality $X \le \lceil \frac{2001}{4} \rceil$.
So if there is a subsequence that satisfies the conditions of the problem, the number $X$ of coins of value 3, must satisfy $X \in \{ \lfloor \frac{2001}{4} \rfloor, \lceil \frac{2001}{4}\rceil \}$. Now in fact there is indeed such a sequence, and furthermore, there is such a sequence where $X$ $=$ $\lfloor \frac{2001}{4} \rfloor$, and there is also such a sequence where  $X=\lceil \frac{2001}{4} \rceil$; consider the infinite sequence 12131213121312131213.... Any contiguous subsequence of this of length 2001 satisfies the conditions of the problem and is a valid arrangement.
