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 $a+b+c+d+e=$ $\qquad$ .
2. 57 .

From five positive integers, if we take any four to find their sum, there are five possible ways, which should result in five sum values. Since the set $\{44,45, 46,47\}$ contains only four elements, there must be two sum values that are equal. Therefore,
$$
\begin{array}{l}
44+44+45+46+47 \\
\leqslant 4(a+b+c+d+e) \\
\leqslant 44+45+46+47+47 \\
\Rightarrow 226 \leqslant 4(a+b+c+d+e) \leqslant 229 .
\end{array}
$$

Since $a, b, c, d, e$ are integers, we get
$$
\begin{array}{l}
4(a+b+c+d+e)=228 \\
\Rightarrow a+b+c+d+e=57
\end{array}
$$