14. In the coordinate plane, points with both coordinates as integers are called integer points, and triangles with all three vertices as integer points are called integer point triangles. Find the number of integer point right triangles $OAB$ with a right angle at the origin $O$ and with $I(2015,7 \times 2015)$ as the incenter.
14. Let point $A$ be in the first quadrant.

Let $\angle x O I=\alpha$.
Then $\tan \alpha=7$,
$k_{O A}=\tan \left(\alpha-\frac{\pi}{4}\right)=\frac{\tan \alpha-1}{1+\tan \alpha}=\frac{3}{4}$
$\Rightarrow k_{O B}=-\frac{4}{3}$.
Since $A$ and $B$ are integer points, set
$$
A\left(4 t_{1}, 3 t_{1}\right), B\left(-3 t_{2}, 4 t_{2}\right)\left(t_{1} 、 t_{2} \in \mathbf{Z}_{+}\right) \text {. }
$$

Thus, $|O A|=5 t_{1},|O B|=5 t_{2}$.
Let the inradius of $\triangle O A B$ be $r$. Then
$r=\frac{\sqrt{2}}{2}|O I|=\frac{\sqrt{2}}{2} \times 5 \sqrt{2} \times 2015=5 \times 2015$.
Also, $r=\frac{|O A|+|O B|-|A B|}{2}$
$\Rightarrow|A B|=|O A|+|O B|-2 r$
$\Rightarrow|A B|^{2}=(|O A|+|O B|-2 r)^{2}=|O A|^{2}+|O B|^{2}$
$\Rightarrow\left(5 t_{1}+5 t_{2}-2 \times 5 \times 2015\right)^{2}=25 t_{1}^{2}+25 t_{2}^{2}$
$\Rightarrow\left(t_{1}+t_{2}-2 \times 2015\right)^{2}=t_{1}^{2}+t_{2}^{2}$.
Set $t_{1}=x+2015, t_{2}=y+2015$. Then
$(x+y)^{2}=(x+2015)^{2}+(y+2015)^{2}$
$\Rightarrow x y=2015 x+2015 y+2015^{2}$
$\Rightarrow(x-2015)(y-2015)=2 \times 2015^{2}$
$=2 \times 5^{2} \times 13^{2} \times 31^{2}$.

Since $|O A|>2 r,|O B|>2 r$, we know
$x-2015, y-2015 \in \mathbf{Z}_{+}$.
Note that, $2 \times 5^{2} \times 13^{2} \times 31^{2}$ has $2 \times 3 \times 3 \times 3=54$ positive divisors.
Thus, there are 54 pairs of $(x, y)$ that satisfy the conditions.
Therefore, there are 54 triangles that satisfy the conditions.