Three. (25 points) Given that $a$ is a positive integer not greater than 2005, and $b, c$ are integers, the parabola $y=a x^{2}+b x+c$ is above the $x$-axis and passes through points $A(-1,4 a+7)$ and $B(3,4 a-1)$. Find the minimum value of $a-b+c$.

The parabola $y=a x^{2}+b x+c$ passes through points $A(-1,4 a+7)$ and $B(3,4 a-1)$. 

Substituting these points into the equation of the parabola, we get:
1. For point $A(-1,4 a+7)$:
\[ 4a + 7 = a(-1)^2 + b(-1) + c \]
\[ 4a + 7 = a - b + c \]
\[ 3a + 7 = -b + c \]
\[ c - b = 3a + 7 \quad \text{(1)} \]

2. For point $B(3,4 a-1)$:
\[ 4a - 1 = a(3)^2 + b(3) + c \]
\[ 4a - 1 = 9a + 3b + c \]
\[ -5a - 1 = 3b + c \]
\[ c + 3b = -5a - 1 \quad \text{(2)} \]

We now have the system of linear equations:
\[ c - b = 3a + 7 \quad \text{(1)} \]
\[ c + 3b = -5a - 1 \quad \text{(2)} \]

Subtract equation (1) from equation (2):
\[ (c + 3b) - (c - b) = (-5a - 1) - (3a + 7) \]
\[ 4b = -8a - 8 \]
\[ b = -2a - 2 \]

Substitute \( b = -2a - 2 \) into equation (1):
\[ c - (-2a - 2) = 3a + 7 \]
\[ c + 2a + 2 = 3a + 7 \]
\[ c = a + 5 \]

Now, we need to find \( a - b + c \):
\[ a - b + c = a - (-2a - 2) + (a + 5) \]
\[ a + 2a + 2 + a + 5 \]
\[ 4a + 7 \]

Since \( a \) is a positive integer not greater than 2005, the minimum value of \( 4a + 7 \) occurs when \( a = 1 \):
\[ 4(1) + 7 = 11 \]

Thus, the minimum value of \( a - b + c \) is:
\[ \boxed{11} \]
Three, since $a$ is a positive integer, the parabola opens upwards.
The parabola $y=a x^{2}+b x+c$ is above the $x$-axis, so,
$$
b^{2}-4 a c<0.
$$
Given $b=-2-2a$ and $c=a+5$, we have
$$
b^{2}-4 a c=(-2-2a)^{2}-4a(a+5)=4+8a+4a^{2}-4a^{2}-20a=4-12a<0.
$$
This implies
$$
a>\frac{1}{3}.
$$
Since $a$ is a positive integer no greater than 2005, we have $a \geqslant 1$.
Thus, $b=-2-2a \leqslant-4, -b \geqslant 4, c=a+5 \geqslant 6$.
When $a=1, b=-4, c=6$, the parabola $y=x^{2}-4 x+6$ satisfies the given conditions.
Therefore, the minimum value of $a-b+c$ is 11.