Solution:

Condition on Alex and Dylan holding $a$ and $a + 9.$ Blair and Corey then draw $2$ of the remaining $50$ cards, and Alex and Dylan are teammates exactly when both of those cards are below $a$ (Alex and Dylan are the high team) or both are above $a + 9$ (the low team). There are $a - 1$ cards below and $52 - (a + 9) = 43 - a$ cards above, so $$p(a) = \frac{\binom{a-1}{2} + \binom{43-a}{2}}{\binom{50}{2}}.$$

The numerator is $\frac{(a-1)(a-2) + (43-a)(42-a)}{2} = a^2 - 44a + 904,$ so $p(a) \ge \frac{1}{2}$ becomes $a^2 - 44a + 904 \ge \frac{25 \cdot 49}{2},$ that is, $(a - 22)^2 \ge \frac{385}{2}.$ Since $a$ is an integer, $|a - 22| \ge 14,$ so $a \le 8$ or $a \ge 36.$

The parabola is smallest at the admissible points closest to $a = 22:$ $p(8) = p(36) = \frac{\binom{7}{2} + \binom{35}{2}}{\binom{50}{2}} = \frac{616}{1225} = \frac{88}{175},$ which is indeed at least $\frac{1}{2}.$ Thus $m + n = 88 + 175 = 263.$