2018 AIME I Problem 12

Each residue class modulo $3$ contains exactly six elements of $U,$ and only the numbers of chosen elements $\equiv 1$ and $\equiv 2$ matter

With $a$ chosen elements $\equiv 1$ and $b$ elements $\equiv 2,$ you need $a \equiv b \pmod 3;$ evaluate $\sum \binom{6}{a}\binom{6}{b}$ using Vandermonde’s identity

The set $U$ contains six elements in each residue class modulo $3.$ If $T$ contains $a$ elements $\equiv 1$ and $b$ elements $\equiv 2 \pmod 3,$ then $s(T) \equiv a + 2b \equiv a - b \pmod 3,$ so $s(T)$ is divisible by $3$ exactly when $a \equiv b \pmod 3;$ the six multiples of $3$ may be included freely, contributing a factor $2^6$ to both the favorable and total counts.

By Vandermonde’s identity, the number of ways to choose the $a$s and $b$s with $a - b = 0$ is $\sum_a \binom{6}{a}^2 = \binom{12}{6} = 924;$ with $a - b = \pm 3$ it is $2\sum_a \binom{6}{a}\binom{6}{a - 3} = 2\binom{12}{9} = 440;$ and with $a - b = \pm 6$ it is $2.$ The favorable choices number $924 + 440 + 2 = 1366$ out of $2^{12}.$

The probability is $\frac{1366}{4096} = \frac{683}{2048},$ which is in lowest terms since $683$ is odd. Thus $m = 683.$