2. (10 points) From 4 integers, arbitrarily select 3, find their average. Then find the sum of this average and the remaining 1 number, which can yield 4 numbers: $8$, $12$, $10 \frac{2}{3}$, and $9 \frac{1}{3}$. The sum of the originally given 4 integers is $\qquad$.
【Analysis】According to the problem, let the original 4 integers be $a, b, c, d$, then $\frac{a+b+c}{3}+d=8(1), \frac{a+b+d}{3}+c=12$ (2), $\frac{a+c+d}{3}+b=10 \frac{2}{3}$ (3), $\frac{b+c+d}{3}+a=9 \frac{1}{3}$ (4). We need to find the sum of the original 4 integers based on this.

【Solution】Let the original 4 integers be $a, b, c, d$,
$$
\begin{array}{l}
\frac{a+b+c}{3}+d=8 \\
\frac{a+b+d}{3}+c=12(2), \\
\frac{a+c+d}{3}+b=10 \frac{2}{3}(3), \\
\frac{b+c+d}{3}+a=9 \frac{1}{3}(4), \\
(1)+(2)+(3)+(4), \text { we get } \\
2(a+b+c+d)=8+12+10 \frac{2}{3}+9 \frac{1}{3},
\end{array}
$$

So, $a+b+c+d=20$,
Therefore, the sum of the original 4 integers is 20.
The answer is: 20.