3. $10,11,11,12,13$.

From these five positive integers, if we take any four numbers and find their sum, we can get five sum values. Since there are only four different sum values in this problem, it indicates that two of the sum values must be equal.
The sum of these five sum values is
$$
4\left(x_{1}+x_{2}+x_{3}+x_{4}+x_{5}\right) \text {. }
$$

Thus, $226=44+44+45+46+47$
$\leqslant 4\left(x_{1}+x_{2}+x_{3}+x_{4}+x_{5}\right)$
$$
\leqslant 44+45+46+47+47=229 \text {. }
$$

Solving this, we get $x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=57$.
Therefore, four of the five numbers are
$$
\begin{array}{l}
57-44=13,57-45=12, \\
57-46=11,57-47=10 .
\end{array}
$$

Thus, the remaining number is
$$
57-(10+11+12+13)=11 \text {. }
$$

Hence, the five numbers are $10, 11, 11, 12, 13$.
3. Given $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ are positive integers. The set of sums obtained by taking any four of these numbers is $\{44, 45, 46, 47\}$. Then these five numbers are $\qquad$ .