Let's divide this sequence into a series of groups. Each group starts with $1$ and is followed by as many $2$s as possible: $$\{1, 2\}, \quad\{1, 2, 2, 2\}, \quad\{1, 2, 2, 2, 2, 2\},\quad \{1, \cdots$$ It is clear that the $k^{th}$ group contains $2k$ terms whose sum equals $1+(2k-1)\times 2= 4k-1$. The numbers of terms in these groups form an arithmetic sequence. The total number of terms in first $n$ groups equal $$2\times 1 + 2\times 2 + \cdots + 2\times n = 2\times\frac{n(n+1)}{2}=n\times (n+1)$$ $$n\times (n+1) \le 2014 \implies n = 44$$ This means that $2014$ terms in the original sequence cover $44$ complete group and $1$ partial group. Hence there are $45$ $one$s in the first 2014 terms and $(2014-45)$ $two$'s. Their sum equal $$45\times 1 + (2014-45)\times 2 = \boxed{3983}$$