Question 54: Find all real numbers $N$ such that there exists a strictly increasing sequence of positive integers $\left\{a_{n}\right\}_{n=1}^{+\infty}$, such that for any positive integer $n$, we have $\frac{a_{2 n}-1+a_{2 n}}{a_{n}}=N$.

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Question 54:
Solution: (1) First, prove that $N$ is a positive integer, and $N \geq 4$.
Since $N=\frac{a_{2n-1}+a_{2n}}{a_{n}}$, it is clear that $N \in Q^{+}$.
First, prove that $N \in \mathrm{Z}^{+}$. Otherwise, let $\mathrm{N}=\frac{\mathrm{p}}{\mathrm{q}}$ be a reduced fraction. Thus, all $\mathrm{a}_{\mathrm{n}}$ are multiples of $\mathrm{q}$.
After dividing each term in the sequence $\left\{\mathrm{a}_{\mathrm{n}}\right\}$ by $\mathrm{q}$, the constant $N$ remains unchanged. Repeating this operation, we find that each $\mathrm{a}_{\mathrm{n}}$ can be divided by any power of $\mathrm{q}$. Therefore, it must be that $\mathrm{q}=1$, i.e., $\mathrm{N} \in \mathrm{Z}^{+}$.
Since $N=\frac{a_{1}+a_{2}}{a_{1}}>2$. We only need to prove that $N \neq 3$.
Otherwise, let $b_{n}=a_{n+1}-a_{n}$, then $\left\{b_{n}\right\}$ is a sequence of positive integers.
$$
\left\{\begin{aligned}
3 a_{n} & =a_{2 n-1}+a_{2 n} \\
3 a_{n+1} & =a_{2 n+1}+a_{2 n+2}
\end{aligned}\right.
$$

Subtracting the two equations, we get:
$$
3 b_{n}=\left(a_{2 n+2}-a_{2 n+1}\right)+2\left(a_{2 n+1}-a_{2 n}\right)+\left(a_{2 n}-a_{2 n-1}\right)=b_{2 n+1}+2 b_{2 n}+b_{2 n-1}
$$

The above equation shows that among $\mathrm{b}_{2 \mathrm{n}+1}, \mathrm{~b}_{2 \mathrm{~m}}, \mathrm{~b}_{2 \mathrm{n}-1}$, there must be one term less than $\mathrm{b}_{\mathrm{n}}$, thus the sequence $\left\{b_{n}\right\}$ has a strictly decreasing infinite subsequence, which is impossible.
Therefore, $N$ is a positive integer not less than 4.
(2) Next, prove that when $N$ is a positive integer not less than 4, there exists a sequence that satisfies the conditions.
According to the conditions, $\frac{a_{1}+a_{2}}{a_{1}}=N$. For simplicity, take $a_{1}=1, a_{2}=N-1$.
Also, $\mathrm{a}_{2 \mathrm{n}-1}+\mathrm{a}_{2 \mathrm{n}}=N \mathrm{a}_{\mathrm{n}}=\left(\left\lceil\frac{N \mathrm{a}_{\mathrm{n}}}{2}\right\rceil-1\right)+\left(\left\lceil\frac{\mathrm{Na}_{n}}{2}\right]+1\right)$, we take $\mathrm{a}_{2 \mathrm{n}-1}=\left\lceil\frac{N a_{n}}{2}\right\rceil-1, \mathrm{a}_{2 \mathrm{n}}=\left[\frac{\mathrm{Na}_{n}}{2}\right]+1$ (for any integer $\mathrm{n} \geq 2$). We will now prove that the sequence constructed in this way satisfies the conditions.
(1) When $\mathrm{N}=4$, the sequence constructed above is $\mathrm{a}_{\mathrm{n}}=2 \mathrm{n}-1\left(n \in Z^{+}\right)$, which clearly satisfies the conditions. In fact, for any positive integer $n$, we have $\frac{(4 n-3)+(4 n-1)}{2 n-1}=4$.
(2) For $N \geq 5$. Take $a_{1}=1, a_{2}=N-1$; when $n \geq 2$, take $a_{2 n-1}=\left\lceil\frac{N a_{n}}{2}\right\rceil-1, a_{2 n}=\left[\frac{N a_{n}}{2}\right]+1$. Then $\frac{a_{2 n-1}+a_{2 n} n}{a_{n}}=\frac{\left.\left|\frac{N a_{n}}{2}\right|+\left\lvert\, \frac{N a_{0} n}{2}\right.\right]}{a_{n}}=N$. And it is clear that $a_{2 n-1}a_{2}$.
Assume that for subscripts $\mathrm{m}$ not greater than $2 \mathrm{n}$, we have $\mathrm{a}_{\mathrm{m}-1}0
$$

Thus, $\mathrm{a}_{2 \mathrm{n}}<\mathrm{a}_{2 \mathrm{n}+1}$.
By mathematical induction, it is known that the sequence $\left\{a_{n}\right\}(n \geq 1)$ constructed above is a strictly increasing sequence of positive integers.