Note that
$$a_{i+1}+1=2\left(a_{i}+1\right) \text{ or } a_{i+1}+1=\frac{a_{i}+a_{i}+2}{a_{i}+2}=\frac{2\left(a_{i}+1\right)}{a_{i}+2} .$$
Hence
$$\frac{1}{a_{i+1}+1}=\frac{1}{2} \cdot \frac{1}{a_{i}+1} \text{ or } \frac{1}{a_{i+1}+1}=\frac{a_{i}+2}{2\left(a_{i}+1\right)}=\frac{1}{2} \cdot \frac{1}{a_{i}+1}+\frac{1}{2} .$$
Therefore,
$$\begin{equation*} \frac{1}{a_{k}+1}=\frac{1}{2^{k}} \cdot \frac{1}{a_{0}+1}+\sum_{i=1}^{k} \frac{\varepsilon_{i}}{2^{k-i+1}} \tag{1} \end{equation*}$$
where $\varepsilon_{i}=0$ or $1$.
Multiplying both sides by $2^{k}\left(a_{k}+1\right)$ and putting $a_{k}=2014$, we get
$$2^{k}=\frac{2015}{a_{0}+1}+2015 \cdot\left(\sum_{i=1}^{k} \varepsilon_{i} \cdot 2^{i-1}\right)$$
where $\varepsilon_{i}=0$ or $1$.
Since $\operatorname{gcd}(2,2015)=1$, we have $a_{0}+1=2015$ and $a_{0}=2014$. Therefore,
$$2^{k}-1=2015 \cdot\left(\sum_{i=1}^{k} \varepsilon_{i} \cdot 2^{i-1}\right)$$
where $\varepsilon_{i}=0$ or $1$.
We now need to find the smallest $k$ such that $2015 \mid 2^{k}-1$. Since $2015= 5 \cdot 13 \cdot 31$, from the Fermat little theorem we obtain $5\mid 2^{4}-1$, $13\mid 2^{12}-1$ and $31 \mid 2^{30}-1$. We also have $\operatorname{lcm}[4,12,30]=60$, hence $5\mid 2^{60}-1$, $13\mid 2^{60}-1$ and $31 \mid 2^{60}-1$, which gives $2015 \mid 2^{60}-1$.
But $5 \nmid 2^{30}-1$ and so $k=60$ is the smallest positive integer such that $2015 \mid 2^{k}-1$. To conclude, the smallest positive integer $k$ such that $a_{k}=2014$ is when $k=60$.

Alternative solution 1:
Clearly all members of the sequence are positive rational numbers. For each positive integer $i$, we have $a_{i}=\frac{a_{i+1}-1}{2}$ or $a_{i}=\frac{2 a_{i+1}}{1-a_{i+1}}$. Since $a_{i}>0$ we deduce that
$$a_{i}= \begin{cases}\frac{a_{i+1}-1}{2} & \text{ if } a_{i+1}>1 \\ \frac{2 a_{i+1}}{1-a_{i+1}} & \text{ if } a_{i+1}<1\end{cases}$$
Thus $a_{i}$ is uniquely determined from $a_{i+1}$. Hence starting from $a_{k}=2014$, we simply run the sequence backwards until we reach a positive integer. We compute as follows.
$$\begin{aligned} & \frac{2014}{1}, \frac{2013}{2}, \frac{2011}{4}, \frac{2007}{8}, \frac{1999}{16}, \frac{1983}{32}, \frac{1951}{64}, \frac{1887}{128}, \frac{1759}{256}, \frac{1503}{512}, \frac{991}{1024}, \frac{1982}{33}, \frac{1949}{66}, \frac{1883}{132}, \frac{1751}{264}, \frac{1487}{528}, \frac{959}{1056}, \frac{1918}{97}, \frac{1821}{194}, \frac{1627}{388}, \\ & \frac{1239}{776}, \frac{463}{1552}, \frac{926}{1089}, \frac{1852}{163}, \frac{1689}{326}, \frac{1363}{652}, \frac{711}{1304}, \frac{1422}{593}, \frac{829}{1186}, \frac{1658}{357}, \frac{1301}{714}, \frac{587}{1428}, \frac{1174}{841}, \frac{333}{1682}, \frac{666}{1349}, \frac{1332}{683}, \frac{649}{1366}, \frac{1298}{717}, \frac{581}{1434}, \frac{1162}{853}, \\ & \frac{309}{1706}, \frac{618}{1397}, \frac{1236}{779}, \frac{457}{1558}, \frac{914}{1101}, \frac{1828}{187}, \frac{1641}{374}, \frac{1267}{748}, \frac{519}{1496}, \frac{1038}{977}, \frac{61}{1954}, \frac{122}{1893}, \frac{244}{1771}, \frac{488}{1527}, \frac{976}{1039}, \frac{1952}{63}, \frac{1889}{126}, \frac{1763}{252}, \frac{1511}{504}, \frac{1007}{1008}, \frac{2014}{1} . \end{aligned}$$
There are 61 terms in the above list. Thus $k=60$.

Alternative solution 2:
Start with $a_{k}=\frac{m_{0}}{n_{0}}$ where $m_{0}=2014$ and $n_{0}=1$ as in alternative solution 1. By inverting the sequence as in alternative solution 1, we have $a_{k-i}=\frac{m_{i}}{n_{i}}$ for $i \geq 0$ where
$$\left(m_{i+1}, n_{i+1}\right)= \begin{cases}\left(m_{i}-n_{i}, 2 n_{i}\right) & \text{ if } m_{i}>n_{i} \\ \left(2 m_{i}, n_{i}-m_{i}\right) & \text{ if } m_{i}<n_{i}\end{cases}$$
Easy inductions show that $m_{i}+n_{i}=2015$, $1 \leq m_{i}, n_{i} \leq 2014$ and $\operatorname{gcd}\left(m_{i}, n_{i}\right)=1$ for $i \geq 0$. Since $a_{0} \in \mathbb{N}^{+}$ and $\operatorname{gcd}\left(m_{k}, n_{k}\right)=1$, we require $n_{k}=1$. An easy induction shows that $\left(m_{i}, n_{i}\right) \equiv\left(-2^{i}, 2^{i}\right)(\bmod 2015)$ for $i=0,1, \ldots, k$.
Thus $2^{k} \equiv 1(\bmod 2015)$. As in the official solution, the smallest such $k$ is $k=60$. This yields $n_{k} \equiv 1(\bmod 2015)$. But since $1 \leq n_{k}, m_{k} \leq 2014$, it follows that $a_{0}$ is an integer.