(1) Discuss the number of roots of the equation with respect to $x$
$$
|x+1|+|x+2|+|x+3|=a
$$

(2) Let $a_{1}, a_{2}, \cdots, a_{n}$ be an arithmetic sequence, and
$$
\begin{array}{l}
\left|a_{1}\right|+\left|a_{2}\right|+\cdots+\left|a_{n}\right| \\
=\left|a_{1}+1\right|+\left|a_{2}+1\right|+\cdots+\left|a_{n}+1\right| \\
=\left|a_{1}-2\right|+\left|a_{2}-2\right|+\cdots+\left|a_{n}-2\right|=507 .
\end{array}
$$

Find the maximum value of the number of terms $n$. (Lin Chang)
(1) According to the graph of the function $y=|x+1|+|x+2|+|x+3|$ (as shown in Figure 3), we can see that:

When $a2$, the equation has two solutions.
(2) Since the equation $|x|=|x+1|=|x-2|$ has no solution, it follows that $n \geqslant 2$ and the common difference is not 0. Let's assume the terms of the sequence are $a-k d(1 \leqslant k \leqslant n, d>0)$. Construct the function:
$$
f(x)=\sum_{k=1}^{n}|x-k d| .
$$

Thus, the condition of the problem is equivalent to $f(x)=507$ having at least three different solutions $a, a+1, a-2$. This condition is also equivalent to the graph of $y=f(x)$ intersecting the horizontal line $y=507$ at least three different points.

Since the graph of $y=f(x)$ is a downward convex broken line with $n+1$ segments symmetric about the line $x=\frac{(n+1) d}{2}$, it intersects the horizontal line $l$ at three points if and only if the broken line has a horizontal segment on $l$, which occurs if and only if $n=2 m$ and $a, a+1, a-2 \in[m d, (m+1) d]$, $f(m d)=507$, i.e., $d \geqslant 3$ and $m^{2} d=507$.
This gives $m^{2} \leqslant \frac{507}{3}, m \leqslant 13$.

Clearly, when $m=13$, taking $d=3, a=41$ satisfies the conditions of the problem.
Therefore, the maximum value of $n$ is 26.