5. Let $x_{1}, x_{2}, \cdots, x_{7}$ be natural numbers, and $x_{1}<x_{2}$ $<\cdots<x_{6}<x_{7}$, also $x_{1}+x_{2}+\cdots+x_{7}=159$. Then the maximum value of $x_{1}+x_{2}+x_{3}$ is $\qquad$
5. 61.

Given that $x_{7} \geqslant x_{6}+1 \geqslant x_{5}+2 \geqslant x_{4}+3 \geqslant x_{3}+4 \geqslant$ $x_{2}+5 \geqslant x_{1}+6$.

Similarly, $x_{6} \geqslant x_{1}+5, x_{5} \geqslant x_{1}+4, x_{4} \geqslant x_{1}+3, x_{3} \geqslant x_{1}+2, x_{2} \geqslant x_{1}+1$.
$$
\begin{array}{l}
\therefore 159=x_{1}+x_{2}+\cdots+x_{7} \geqslant 7 x_{1}+(1+2+\cdots+6) \\
\quad=7 x_{1}+21 . \\
\therefore x_{1} \leqslant \frac{138}{7}<20 .
\end{array}
$$

Therefore, the maximum value of $x_{1}$ is 19.
Taking $x_{1}=19$, then $140=x_{2}+x_{3}+\cdots+x_{7} \geqslant 6 x_{2}+(1+2+\cdots+5)$. Thus, $x_{2} \leqslant 20$.

Taking $x_{1}=19, x_{2}=20$, then $120=x_{3}+\cdots+x_{7} \geqslant 5 x_{3}+(1+2+3+4)$. Thus, $x_{3} \leqslant 22$.

Following this, we can get the 7 numbers that meet the conditions: $19,20,22,23,24,25,26$. Therefore, the maximum value sought is 61.