$$
\begin{array}{l}
\text { 8. Let } \\
\begin{array}{l}
A=\{1,2, \cdots, 2014\}, \\
B_{i}=\left\{x_{i}, y_{i}\right\}(i=1,2, \cdots, t)
\end{array}
\end{array}
$$

be $t$ pairwise disjoint binary subsets of $A$, and satisfy the conditions
$$
\begin{array}{l}
x_{i}+y_{i} \leqslant 2014(i=1,2, \cdots, t), \\
x_{i}+y_{i} \neq x_{j}+y_{j}(1 \leqslant i<j \leqslant t) .
\end{array}
$$

Then the maximum value of $t$ is . $\qquad$
8. 805 .

On the one hand, for any $1 \leqslant i<j \leqslant t$, we have
$$
\left\{x_{i}, y_{i}\right\} \cap\left\{x_{j}, y_{j}\right\}=\varnothing \text {. }
$$

Thus, $x_{1}, x_{2}, \cdots, x_{t}, y_{1}, y_{2}, \cdots, y_{t}$ are $2 t$ distinct integers. Therefore,
$$
\sum_{i=1}^{t}\left(x_{i}+y_{i}\right) \geqslant \sum_{i=1}^{2 t} i=t(2 t+1) .
$$

On the other hand, for any $1 \leqslant i<j \leqslant t$, we have
$$
\begin{array}{l}
x_{i}+y_{i} \neq x_{j}+y_{j}, \text { and } x_{i}+y_{i} \leqslant 2014, \\
\text { then } \sum_{i=1}^{t}\left(x_{i}+y_{i}\right) \\
\leqslant \sum_{i=1}^{i}(2015-i)=\frac{t(4029-t)}{2} .
\end{array}
$$

From equations (1) and (2), we get $t \leqslant 805$.
Below are two examples showing that $t=805$ meets the requirements.
$$
\begin{array}{l}
\text { (1) }\{k, 1208+k\}(k=1,2, \cdots, 402), \\
\{k, 403+k\}(k=403,404, \cdots, 805) ; \\
\text { (2) }\{k, 1611-2 k\}(k=1,2, \cdots, 402), \\
\{k, 2016-k\}(k=403,404, \cdots, 805) .
\end{array}
$$