8. If the sum of $n$ consecutive positive integers equals 2014, then the maximum value of $n$ is $\qquad$ .
8.53.

Let these $n$ consecutive positive integers be
$$
k, k+1, \cdots, k+n-1(k \geqslant 1) .
$$

Then $n k+\frac{n(n-1)}{2}=2014$
$$
\Rightarrow n(2 k+n-1)=4028 \text {. }
$$

Thus, $n$ and $2 k+n-1$ are both positive divisors of 4028.
Since $n<2 k+n-1$, then
$$
n \leqslant[\sqrt{4028}]=63 \text {. }
$$

Note that, $4028=2^{2} \times 19 \times 53$, its largest divisor not greater than 63 is 53.

Therefore, $n=53, k=12$, at this point, the sum of 53 consecutive positive integers $12,13, \cdots, 64$ is exactly 2014.