385 A certain school's 2014 graduates numbered 2014 students. The school's six leaders must sign each student's graduation album. It is known that each leader must and can only use one of the three designated colored pens, and the color choices for the six leaders signing 2014 albums can be represented as a $6 \times 2014$ grid, called a color selection scheme. Find the smallest positive integer $n$, such that there exists a color selection scheme where in any $n$ albums, there are 8 albums where any two albums have at most four leaders' signatures in the same color.
Let 1, 2, 3 represent three colors, and the colors used by six school leaders for signing in each memorial album be represented by a six-element ordered array
$$
\left(x_{1}, x_{2}, \cdots, x_{6}\right)\left(x_{i} \in\{1,2,3\}\right)
$$

This array is referred to as the sequence group $\left(x_{1}, x_{2}, \cdots, x_{6}\right)$. For $x_{2}, x_{3}, \cdots, x_{6} \in\{1,2,3\}$, the set
$$
\left\{\left(1, x_{2}, \cdots, x_{6}\right),\left(2, x_{2}, \cdots, x_{6}\right),\left(3, x_{2}, \cdots, x_{6}\right)\right\}
$$

is called a Type 1 set. Clearly, all sequence groups $\left(x_{1}, x_{2}, \cdots, x_{6}\right)\left(x_{i} \in\{1,2,3\}\right)$ belong to 243 types.

Given $2014=8 \times 243+70$, we know that among 2014 memorial albums, there are nine albums of the same type. Select these nine albums;

Given $2005=8 \times 243+61$, we know that among the remaining 2005 albums, there are another nine albums of the same type. Select these nine albums;

Continuing this process, we can select seven times, obtaining a total of 63 albums. Among these, each set of eight albums includes at least two albums selected in the same round, which must be of the same type. This means that at least five school leaders signed with the same color in these two albums. This indicates that for any coloring scheme, there are 63 albums that do not meet the requirement. Therefore, $n \geqslant 64$.

On the other hand, consider the set
$$
S=\left\{\left(x_{1}, x_{2}, \cdots, x_{6}\right) \mid x_{i} \in\left\{1,2,3 \mid, \sum_{i=1}^{6} x_{i}=0(\bmod 3)\right\}\right. \text {. }
$$

The number of elements in set $S$ is 243, and any two different elements in set $S$ have at most four components the same. Arbitrarily remove 19 elements from set $S$, leaving 224 elements to form set $T$. Divide the 2014 memorial albums into 224 groups, with the first 223 groups each containing 9 albums, and the last group containing 7 albums. Ensure that each group corresponds to an element in set $T$ (i.e., the signatures in the same group are identical, while those in different groups are different). Clearly, any 64 albums must belong to at least eight groups, thus there exist eight albums corresponding to eight different elements in set $T$. Therefore, any two albums among these have at most four school leaders' signatures in the same color.

In summary, the minimum value of $n$ is 64.
（Fang Tinggang, Chengdu No. 7 High School, Sichuan, 610041）