$2 \cdot 89$ Let $s$ be a list of positive integers (the list can contain the same number multiple times), which includes the number 68. The arithmetic mean of the numbers in $s$ is 56, but if 68 is removed, the arithmetic mean of the remaining numbers drops to 55. What is the largest number that could appear in $s$?
[Solution] Let $n$ be the number of positive integers in $s$, and its $n$ numbers are $a_{1}, a_{2}, \cdots, a_{n-1}$, 68. According to the problem, we have
$$
\begin{array}{l}
\frac{a_{1}+a_{2}+a_{3}+\cdots+a_{n-1}+68}{n}=56 \\
\frac{a_{1}+a_{2}+\cdots+a_{n-1}}{n-1}=55
\end{array}
$$

From (1) and (2), we get $56 n-68=55(n-1)$
Therefore,
$$
n=13 \text {. }
$$

Then $a_{1}+a_{2}+\cdots+a_{12}=55 \times 12=660$.
To make $a_{1}, a_{2}, \cdots, a_{12}$ contain the largest positive integer, the other 11 numbers should be as small as possible. Therefore, among these 12 numbers, 11 are taken as 1, then the other number
$$
660-11=649
$$

is the largest number that may appear in $s$.