2・76 Let $S=\{1,2,3,4\}, n$ terms of the sequence: $a_{1}, a_{2}, \cdots, a_{n}$ have the following property: for any non-empty subset $B$ of $S$ (the number of elements in $B$ is denoted as $|B|$), there are adjacent $|B|$ terms in the sequence that exactly form the set $B$. Find the minimum value of $n$.
2. 77 In a $100 \times 25$ rectangular table, each cell is filled with a non-negative real number, the number filled in the $i$-th row and $j$-th column is $x_{i j}(i=1,2, \cdots, 100 ; j=1,2, \cdots, 25)$ (as shown in Table 1). However, the numbers in each column of Table 1 are rearranged in descending order from top to bottom as $x_{1, j}^{\prime} \geqslant x_{2 j}^{\prime} \geqslant \cdots \geqslant x_{100 j}^{\prime}(j=1,2, \cdots, 25)$
\begin{tabular}{|c|c|c|c|}
\hline$x_{1,1}$ & $x_{1,2}$ & $\cdots$ & $x_{1,25}$ \\
\hline$x_{2,1}$ & $x_{2,2}$ & $\cdots$ & $x_{2,25}$ \\
\hline$\cdots$ & $\cdots$ & $\cdots$ & $\cdots$ \\
\hline$x_{100,1}$ & $x_{100,2}$ & $\cdots$ & $x_{100.25}$ \\
\hline
\end{tabular}

Table 1
\begin{tabular}{|c|c|c|c|}
\hline$x_{1,1}^{\prime}$ & $x_{1,2}^{\prime}$ & $\cdots$ & $x_{1,25}^{\prime}$ \\
\hline$x_{2,1}^{\prime}$ & $x_{2,2}^{\prime}$ & $\cdots$ & $x_{2,25}^{\prime}$ \\
\hline$\cdots$ & $\cdots$ & $\cdots$ & $\cdots$ \\
\hline$x_{100,1}^{\prime}$ & $x_{100,2}^{\prime}$ & $\cdots$ & $x_{100,25}^{\prime}$ \\
\hline
\end{tabular}

Table 2
Find the smallest natural number $k$, such that if the numbers filled in Table 1 satisfy $\sum_{j=1}^{25} x_{i j} \leqslant 1$, $(i=1,2, \cdots, 100)$, then when $i \geqslant k$, in Table 2, it can be guaranteed that
$$
\sum_{j=1}^{25} x_{i j} \leqslant 1
$$

holds.
[Solution] The minimum value of $n$ is 8.
First, we prove that each number in $s$ appears at least twice in the sequence $a_{1}, a_{2}, \cdots, a_{n}$.

In fact, if a number in $s$ appears only once in this sequence, since there are 3 subsets containing this number, but in the sequence, there are at most two pairs of adjacent terms containing this number, it is impossible for all 3 subsets containing this number to appear as adjacent terms in the sequence. This is a contradiction.
Therefore, $n \geqslant 8$.
On the other hand, the sequence $3,1,2,3,4,1,2,4$ satisfies the problem's conditions and has only 8 terms. Thus, the minimum value of $n$ is 8.
[Solution] The minimum value of $k$ is 97.
In fact, if we take
$$
x_{ij}=\left\{\begin{array}{l}
0,4, \quad (j-1)+1 \leqslant i \leqslant 4j \\
\frac{1}{24}, \quad \text{otherwise} \quad (i=1,2, \cdots, 100)
\end{array} \quad (j=1,2, \cdots, 25)\right.
$$

Then, we have
$$
\sum_{j=1}^{25} x_{ij}=0+24 \times \frac{1}{24}=1 \quad (i=1,2, \cdots, 100)
$$

which satisfies the problem's conditions. After rearrangement, we have
$$
x_{ij}^{\prime}=\left\{\begin{array}{l}
\frac{1}{24}, \quad 1 \leqslant i \leqslant 96, \\
0, \quad 97 \leqslant i \leqslant 100 .
\end{array} \quad (j=1,2, \cdots, 25)\right.
$$

At this point,
$$
\sum_{j=1}^{25} x_{ij}^{\prime}=25 \times \frac{1}{24}>1, \quad (1 \leqslant i \leqslant 96) .
$$

Thus, the minimum value of $k \geqslant 97$.
On the other hand, since the last three rows of Table 2 contain only 75 numbers, and Table 1 has 100 rows, there must be at least one row in Table 1 where all its elements appear in the first 97 rows of Table 2. Suppose the $r$-th row of Table 1 has all its numbers
$$
x_{r,1}, x_{r,2}, \cdots, x_{r,25}
$$

appearing in the first 97 rows of Table 2. Therefore, when $i \geqslant 97$, we have
$$
\begin{array}{l}
x_{ij}^{\prime} \leqslant x_{rj}, \quad (j=1,2, \cdots, 25) \\
\sum_{j=1}^{25} x_{ij}^{\prime} \leqslant \sum_{j=1}^{25} x_{rj} \leqslant 1 .
\end{array}
$$

In summary, the minimum value of $k$ is 97.