It is easy to see that $ 4$ is not good (without using proporties of rational and irrational numbers, but testing possibilities). If $ ABCDE$ is a pentagon with integer sides but $ ACEBD$ is a star with irrational sides, the conditions to $ n = 5$ be a good number are fulfilled. We can thus take $ P_1 = (0, 0)$ $ P_2 = (5, 12)$ $ P_3 = (9, 9)$ $ P_4 = (15, 1)$ $ P_5 = (15, 0)$

Solution of part (1) :- It is easy to check that $n=3$ and $n=4$ does not work. For $n=5$ we may take five points $(0,0),(1,0),(0,7),(5,3),(8,7)$ which clearly works. So minimum good number is $\boxed 5$

Here's a potential solution for Part B: Consider the sets $A = { (1, 0), (2, 0), ..., (669, 0) }$ , $B={(1,1), (2, 1), ..., (668, 1)}$, $C={(1, 2), (2, 2), ..., (668, 2)}$. We claim that the $2005$ points in the set $S_{2005} = A\cup B\cup C$ satisfies the conditions of the problem. Consider a pair of distinct points $P_i = (x_i , y_i)$ and $P_j = (x_j , y_j)$ in the set $S_{2005}$. If either $x_i =x_j$ or $y_i = y_j$ , then clearly $P_iP_j$ is rational. Otherwise, either $|x_i-x_j|= 1$ or $2$, or $|y_i-y_j|= 1$ or 2, or both. Hence $|P_iP_j|^2$ can be written in the form of either $n^2+1$ or $n^2 + 4$ for some positive integer $n$. For a pair of positive integers $n$ and $m$ with $n > m$, we have $n^2 - m^2 > n^2\geq n^2- (n- 1)^2 = 2n - 1$. It follows that for a positive integer $n$ , $n^2 +1$ and $n^2 +4$ are not squares of an integer. We conclude that $(P_i , P_j)$ is rational iff line $P_iP_j$ is parallel to one of the axes. If $P_iP_j$ is rational with $y_i = y_j$, then we set $P_k = (X_k , Y_k)$, with $x_k\neq x_i$ , $x_k\neq x_j$, and $y_k\neq y_i$. (We can do so because there are $668$ distinct $x$ values and three distinct $y$ values. ). Then $(P_i, P_j , P_k)$ is good. In exactly the same way, we can deal with the case when $P_iP_j$ is rational with $x_i = x_j$. If $P_iP_j$ is irrational, then $x_i\neq x_j$ and $y_i\neq yj$. We set $P_k = (x_i,y_j)$. Then $(P_i , P_j , P_k)$ is good. Therefore, set $S_{2005}$ satisfies the conditions of the problem and $n=2005$ is good $\blacksquare$ Note: Let $S_6 = S_5\cup ({P_6 = (-24, 0)})$ and $S_7 = S_6\cup ({P7 = (- 24, 7) })$. We can now see both $6$ and $7$ are good. For every positive integer $n\geq 8 $, it is possible to generalize the construction shown above to show that $n$ is good. Therefore, all integers greater than $4$ are good.