6. Let $a_{1}, a_{2}, \cdots, a_{\mathrm{n}}$ be an arithmetic sequence, and
$$
\sum_{i=1}^{n}\left|a_{i}+j\right|=2028(j=0,1,2,3) \text {. }
$$

Then the maximum value of the number of terms $n$ is
6. 52.

Since the equation $|x|=|x+1|=|x+2|$ has no solution, we have $n \geqslant 2$ and the common difference is not 0.
Assume the general term of the sequence is $a-k d(1 \leqslant k \leqslant n, d>0)$.
Construct the function $f(x)=\sum_{k=1}^{n}|x-k d|$.
The given condition is equivalent to $f(x)=2028$ having at least four distinct roots $a, a+1, a+2, a+3$, which means the graph of $y=f(x)$ intersects the line $l: y=2028$ at least four times.

Notice that the graph of $y=f(x)$ is a downward convex broken line with $n+1$ segments symmetric about the line $y=\frac{(n+1) d}{2}$. It intersects the line $l$ at least four times if and only if there is a horizontal segment of the broken line on the line $l$, which happens if and only if $n=2 m$ and $a, a+1, a+2, a+3 \in[m d,(m+1) d]$, $f(m d)=2028$, i.e., $d \geqslant 3$ and $m^{2} d=2028$.
This gives $m^{2} \leqslant \frac{2028}{3}=676 \Rightarrow m \leqslant 26$.
Clearly, when $m=26$, taking $d=3, a=78$ satisfies the conditions of the problem.
Therefore, the maximum value of $n$ is 52.