Three, (17 points) The quadratic trinomial $x^{2}-x-2 n$ can be factored into the product of two linear factors with integer coefficients.
(1) If $1 \leqslant n \leqslant 30$, and $n$ is an integer, how many such $n$ are there?
(2) When $n \leqslant 2005$, find the largest integer $n$.


(1) Notice that
$$
x^{2}-x-2 n=\left(x-\frac{1+\sqrt{1+8 n}}{2}\right)\left(x-\frac{1-\sqrt{1+8 n}}{2}\right),
$$

then we should have
$$
1+8 n=9,25,49,81,121,169,225,289, \cdots \text {. }
$$

The corresponding solutions for $n$ are $1,3,6,10,15,21,28,36$ (discard the last one).
Therefore, when $1 \leqslant n \leqslant 30$, there are 7 integers $n$ that satisfy the condition.
(2) Observing the sequence $1,3,6,10, \cdots$ and the relationship between its terms, we can deduce that
$$
1=1,3=1+2,6=1+2+3,10=1+2+3+4 \text {. }
$$

Thus, as long as $n=1+2+3+\cdots+k \leqslant 2005$, and $k$ is the largest integer satisfying this inequality, then the sum $n$ is what we are looking for.
From $n=\frac{(1+k) k}{2} \leqslant 2005$, we can solve by observation and trial:
When $k=61$, we get $n=1891$;
When $k=62$, we get $n=1953$;
When $k=63$, we get $n=2016>2005$.
Therefore, the largest integer $n$ is 1953.