2. From five positive integers $a, b, c, d, e$, any four are taken to find their sum, resulting in the set of sums $\{44,45,46,47\}$, then $a+b+c+d+e=$ $\qquad$
Given that from five positive integers $a, b, c, d, e$, any four are summed, resulting in 5 values. According to the problem, there must be two of these values that are equal. Note that the sum of these 5 values is $4(a+b+c+d+e)$, which is a multiple of 4, while $44+45+46+47 \equiv 2(\bmod 4)$. Therefore, the two equal values are 46. So, $4(a+b+c+d+e)=44+45+46+46+47=40 \times 5+28=228$ $\Rightarrow a+b+c+d+e=57$.