2. On some planets, there are $2^{N}(N \geqslant 4)$ countries, each with a flag composed of $N$ unit squares forming a rectangle of width $N$ and height 1. Each unit square is either yellow or blue. No two countries have the same flag. If $N$ flags can be arranged to form an $N \times N$ square such that all $N$ unit squares on the main diagonal have the same color, then this set of $N$ flags is called "perfect." Find the smallest positive integer $M$ such that any $M$ different flags must contain $N$ flags that form a perfect set.
2. Note: In the problem-solving process, the diagonals of the squares involved are all main diagonals.
Let $M_{N}$ be the smallest positive integer that satisfies the condition.
First, prove: $M_{N}>2^{N-2}$.
Consider $2^{N-2}$ flags, where the first unit square of each flag is yellow, and the second unit square is blue. Then, both colors will appear on the diagonal of the $N \times N$ square formed by $N$ flags from these flags.
Next, prove: $M_{N} \leqslant 2^{N-2}+1$.
Use mathematical induction on $N$.
First, consider the case when $N=4$.
Assume there are 5 flags, each composed of four unit squares, divided into two parts, each part consisting of two unit squares, and each flag is denoted as $L R$, where $1 \times 2$ flags $L, R \in S=\{B B, B Y, Y B, Y Y\}$, representing the left and right parts of the flag, respectively.
Directly verify the following conclusions for two $1 \times 2$ flags.
(1) For each $A \in S$, there exists a unique $1 \times 2$ flag $C \in S$ such that $A$ and $C$ cannot form a $2 \times 2$ square with the same color diagonal (if $A$ is $B B$, then $C$ is $Y Y$; if $A$ is $B Y$, then $C$ is $B Y$).
(2) For different $A_{1}, A_{2}, A_{3} \in S$, there exist two that can form a $2 \times 2$ square with a yellow diagonal, and there exist two that can form a $2 \times 2$ square with a blue diagonal (if it does not contain $B B$, then $(B Y, Y B)$ satisfies the conclusion for both cases; if it does not contain $B Y$, then $(Y B, Y Y)$ and $(B B, Y B)$ satisfy the conclusion).

Let the left and right parts of the 5 flags have $l$ and $r$ different types, respectively. Then the total number of flags is $5 \leqslant l r$. Therefore, at least one of $l$ and $r$ is not less than 3 (assume $r \geqslant 3$).

On the other hand, since $l, r \leqslant 4$, there are two flags with the same right part, denoted as $L_{1} R_{1}$ and $L_{2} R_{1} (L_{1} \neq L_{2})$.

Since $r \geqslant 3$, there exist $L_{3} R_{3}$ and $L_{4} R_{4}$ such that $R_{1}, R_{3}, R_{4}$ are pairwise different.

Let $L^{\prime} R^{\prime}$ be the remaining flag. By (1), either $(L^{\prime}, L_{1})$ or $(L^{\prime}, L_{2})$ can form a $2 \times 2$ square with the same color diagonal (assume $L^{\prime}, L_{2}$ form a $2 \times 2$ square with a blue diagonal); by (2), among the flags $L_{1} R_{1}, L_{3} R_{3}, L_{4} R_{4}$, there exist two whose right parts can form a $2 \times 2$ square with a blue diagonal. Thus, these four flags can form a $4 \times 4$ square with all unit squares on the diagonal being blue.
For $N>4$, assume the conclusion holds for $N-1$.
Consider any $2^{N-2}+1$ flags of length $N$, and arrange them into a large flag of size $\left(2^{N-2}+1\right) \times N$. Since the flags are different, there must be a column containing two colors (assume the first column). By the pigeonhole principle, there exists a color that appears at least
$$
\left\lceil\frac{2^{N-2}+1}{2}\right\rceil=2^{N-3}+1
$$
times in this column (assume blue).

Call a flag with the first unit square blue a "good" flag.

Consider all good flags and remove the first unit square from each of these flags, resulting in at least
$$
2^{N-3}+1 \geqslant M_{N-1}
$$
flags of length $N-1$. By the induction hypothesis, among these flags, there exist $N-1$ flags that can form a square $Q$ with the same color diagonal. Now, add back the removed unit squares to form an $(N-1) \times N$ rectangle, with the goal of adding one more flag on top of this rectangle.

If the diagonal of square $Q$ is yellow, any flag with the first unit square yellow (since the first column has two colors and the first unit square is yellow in the flags not used in square $Q$) can be added to this $(N-1) \times N$ rectangle to form a square of size $N \times N$ that meets the condition;

If the diagonal of square $Q$ is blue, since
$$
2^{N-3}+1-(N-1)>0,
$$
any good flag can be added to this $(N-1) \times N$ rectangle to form a square of size $N \times N$ that meets the condition.
Therefore, $M_{N}=2^{N-2}+1$.