5. (10 points) Let $a, b, c, d, e$ all be natural numbers, and $a<b<c<d<e, a+2b+3c+4d+5e=300$, then the maximum value of $a+b$ is $\qquad$ .
【Analysis】If $a, b, c, d, e$ are all consecutive natural numbers, it also fits the problem's context. Assume they are: $1, 2, 3, 4, 5$, then $a+2 b+3 c+4 d+5 e=55$. If each number increases by 1, then the sum of $a+2 b+3 c+4 d+5 e$ increases by 15; then look at $(300 - 55)$ to see how many 15s there are; and then determine the values of $a$ and $b$ based on the remainder.

【Solution】If $a, b, c, d, e$ are all consecutive natural numbers, it also fits the problem's context.
Assume they are: $1, 2, 3, 4, 5$,
then, $a+2 b+3 c+4 d+5 e$
$$
\begin{array}{l}
=1+2 \times 2+3 \times 3+4 \times 4+5 \times 5 \\
=55
\end{array}
$$

If they all increase by 1, then the total sum will increase by:
$$
\begin{array}{l}
1+2+3+4+5=15 \\
(300-55) \div 15=16 \cdots 5
\end{array}
$$

The remainder is 5, which can only be added to $e$,
so, $a=1+16=17, b=2+16=18$,
so, the maximum value of $a+b$ is: $17+18=35$.
Answer: The maximum value of $a+b$ is 35.
Therefore, the answer is: 35.