6. There are $n(n>3)$ integers on a circle with a sum of 94, each of which is equal to the absolute value of the difference between the two numbers that follow it (in a clockwise direction). Then all possible values of $n$ are
6.141 .

Obviously, these $n$ integers are all non-negative.
Let $a$ be the largest one, and the next four numbers be $b, c, d, e$ respectively. Then $a=|b-c|$.
Thus, $b=a$ or $c=a$.
When $b=a$, $c=0$.
From $b=|c-d|$, we know $d=a$.
Also, from $c=|d-e|$, we get $d=e$.
Therefore, $n$ is a multiple of 3.
Assume $n=3m$. Then these $n$ numbers are $a, a, 0, a, a, 0, \cdots, a, a, 0$.
Since their sum is $2ma=94$, we have $m=47, a=1$ or $m=1, a=47$.
Thus, $n=141$ or 3 (not valid).
When $c=a$, $b=0$.
Thus, $d=a, e=0$, and these $n$ numbers are $a, 0, a, a, 0, a, \cdots, a, 0, a$.
Similarly, we get $n=141$ or 3 (not valid). Hence, $n$ can only be 141.