1. Given the sequence of positive integers $\left\{y_{n}\right\}$ satisfies
$$
y_{n+2}=\frac{y_{n}+2013}{1+y_{n+1}}(n \geqslant 1) \text {. }
$$

Find the minimum value of $y_{1}+y_{2}$.
Given by the problem, we have
$$
\begin{array}{l}
y_{n+2}+y_{n+2} y_{n+1}=y_{n}+2013, \\
y_{n+3}+y_{n+3} y_{n+2}=y_{n+1}+2013 .
\end{array}
$$

Subtracting the two equations, we get $y_{n+3}-y_{n+1}=\frac{y_{n+2}-y_{n}}{y_{n+2}+1}$. Let $a_{n}=y_{n+2}-y_{n}$. Then
$$
\begin{array}{l}
\left|a_{n+1}\right|=\left|\frac{a_{n}}{y_{n+2}+1}\right| \leqslant \frac{\left|a_{n}\right|}{2} \\
\Rightarrow\left|a_{n}\right| \leqslant \frac{\left|a_{1}\right|}{2^{n-1}} .
\end{array}
$$

When $n$ is sufficiently large, $a_{n}=0$. Since $a_{n}=\left(y_{n+2}+1\right) a_{n+1}$, it follows that $a_{n}=0 \Rightarrow y_{n+2}=y_{n}$ (for any $n \in \mathbf{Z}_{+}$). Substituting into the given equation, we get
$$
y_{1} y_{2}=2013=3 \times 11 \times 61 \text {. }
$$

Thus, the minimum value of $y_{1}+y_{2}$ is 94.