Suppose that $P$ has coordinates $P(0,2a)$ for some real number $a$. Since $P$ has $y$-coordinate greater than 0 and less than 100, then $0 < 2a < 100$ or $0 < a < 50$. We determine an expression for the radius of the circle in terms of $a$ and then determine how many values of $a$ give an integer radius. We determine the desired expression by first finding the coordinates of the centre, $C$, of the circle in terms of $a$, and then calculating the distance from $C$ to one of the points $O, P, Q$. If a circle passes through the three vertices $O, P$ and $Q$ of a triangle, then its centre is the point of intersection of the perpendicular bisectors of the sides $OP, OQ$, and $PQ$ of the triangle. We determine the centre of the circle by finding the point of intersection of the perpendicular bisectors of $OP$ and $OQ$. (We could use $PQ$ instead, but this would be more complicated algebraically.) Since $O$ has coordinates $(0,0)$ and $P$ has coordinates $(0,2a)$, then $OP$ is vertical so its perpendicular bisector is horizontal. The midpoint of $OP$ is $(\frac{1}{2}(0+0), \frac{1}{2}(0+2a)) = (0, a)$. Therefore, the perpendicular bisector of $OP$ is the horizontal line through $(0, a)$, and so has equation $y = a$. Since $O$ has coordinates $(0,0)$ and $Q$ has coordinates $(4,4)$, then $OQ$ has slope $\frac{4-0}{4-0} = 1$. Therefore, a line perpendicular to $OQ$ has slope -1. The midpoint of $OQ$ is $(\frac{1}{2}(0+4), \frac{1}{2}(0+4)) = (2,2)$. Therefore, the perpendicular bisector of $OQ$ has slope -1 and passes through $(2,2)$, so has equation $y - 2 = (-1)(x - 2)$ or $y = -x + 4$. The centre of the desired circle is thus the point of intersection of the lines with equations $y = a$ and $y = -x + 4$. The $y$-coordinate of this point is $a$ and the $x$-coordinate is obtained by solving $a = -x + 4$ and obtaining $x = 4 - a$. Therefore, the coordinates of $C$ are $(4 - a, a)$. The radius, $r$, of the circle is the distance from $C$ to any of the three points $O, P$ and $Q$. It is easiest to find the distance from $O$ to $C$, which is $r = \sqrt{(4-a)^{2} + a^{2}} = \sqrt{a^{2} - 8a + 16 + a^{2}} = \sqrt{2a^{2} - 8a + 16}$. We rewrite this as $r = \sqrt{2(a^{2} - 4a + 8)} = \sqrt{2(a^{2} - 4a + 4 + 4)} = \sqrt{2((a-2)^{2} + 4)} = \sqrt{2(a-2)^{2} + 8}$. Since $(a-2)^{2} \geq 0$ and $(a-2)^{2} = 0$ only when $a = 2$, then the minimum value of $2(a-2)^{2} + 8$ is 8 and this occurs when $a = 2$. Thus, $r \geq \sqrt{8}$. The expression $\sqrt{2(a-2)^{2} + 8}$ is decreasing from $a = 0$ to $a = 2$ and then increasing from $a = 2$ to $a = 50$. When $a = 0, r = \sqrt{2(a-2)^{2} + 8} = \sqrt{2(-2)^{2} + 8} = 4$. When $a = 2, r = \sqrt{2(a-2)^{2} + 8} = \sqrt{2(0)^{2} + 8} = \sqrt{8} \approx 2.83$. When $a = 50, r = \sqrt{2(a-2)^{2} + 8} = \sqrt{2(48)^{2} + 8} = \sqrt{4616} \approx 67.94$. Therefore, when $0 < a \leq 2$, we have $\sqrt{8} \leq r < 4$ and when $2 \leq a < 50$, we have $\sqrt{8} \leq r < \sqrt{4616}$. The expression $r = \sqrt{2(a-2)^{2} + 8}$ will take every real number value in each of these ranges, because $b = 2(a-2)^{2} + 8$ represents the equation of a parabola which is a 'smooth' curve. Between $\sqrt{8} \approx 2.83$ and 4, there is one integer value (namely, 3) which is achieved by the expression. (We do not count 4 since it is an endpoint that is not included.) Between $\sqrt{8} \approx 2.83$ and $\sqrt{4616} \approx 67.94$, there are 65 integer values (namely, 3 to 67, inclusive) which are achieved by the expression. In total, there are $1 + 65 = 66$ integer values achieved by the expression in the allowable range for $a$, so there are 66 positions of $P$ for which the radius is an integer.