Problem 298

### Solution:

#### Part (1):

Given that $\{a_{n}\}$ is an arithmetic sequence, we can express $a_{n}$ as $a_{n}=an+b$. Knowing that $a_{n+1} > a_{n} > 0$, it implies $a > 0$.

From the condition ${a}_{n}-\frac{1}{{a}_{n}}＜n$, we derive:
$$\begin{align*} {a}_{n}^{2}-1 &＜ n{a}_{n} \\ \Rightarrow \left(an+b\right)^{2}-1 &＜ n\left(an+b\right) \\ \Rightarrow (a^{2}-1)n^{2}+\left(2ab-b\right)n+b^{2}-1 &＜ 0 \quad \text{(1)} \end{align*}$$

From the condition $n＜{a}_{n}+\frac{1}{{a}_{n}}$, we derive:
$$\begin{align*} {a}_{n}^{2}+1 &＞ n{a}_{n} \\ \Rightarrow \left(an+b\right)^{2}+1 &＞ n\left(an+b\right) \\ \Rightarrow (a^{2}-1)n^{2}+\left(2ab-b\right)n+b^{2}+1 &＞ 0 \quad \text{(2)} \end{align*}$$

For inequalities (1) and (2) to hold for all $n \in N^{*}$, we must have:
$$\left\{ \begin{array}{l} {a}^{2}-1=0 \\ 2ab-b=0 \end{array} \right.$$
Solving this system of equations gives us:
$$\left\{ \begin{array}{l} a=1 \\ b=0 \end{array} \right.$$
Substituting these values back verifies that the inequalities hold true.

Therefore, we have $a_{n}=n$, and the sum of its first $n$ terms is:
$${S}_{n}=\frac{(1+n)n}{2}=\frac{1}{2}{n}^{2}+\frac{1}{2}n$$
Encapsulating the final answer, we get:
$$\boxed{{S}_{n}=\frac{1}{2}{n}^{2}+\frac{1}{2}n}$$

#### Part (2):

Given the conditions of the sequence $\{a_{n}\}$ and the existence of $m \geqslant 2$ such that $a_{mn}=ma_{n}$ for any $n \in N^{*}$, we analyze the implications.

From the given inequalities, we have:
$$-1 ＜ a_{n}(a_{n}-n) ＜ 1$$
for all $n \in N^{*}$.

Let $n=m^{k}q$ where $m$, $k$, $q \in N^{*}$. Then, we have:
$$-\frac{1}{{m}^{2k}} ＜ a_{q}(a_{q}-q) ＜ \frac{1}{{m}^{2k}}$$
for all $q \in N^{*}$.

As $k \rightarrow +\infty$, it implies $a_{q}-q=0$, hence $a_{q}=q$ for all $q \in N^{*}$.

This leads to $a_{q}-a_{q-1}=1$, indicating that the sequence $\{a_{n}\}$ is an arithmetic sequence.

Encapsulating the final conclusion, we get:
$$\boxed{\text{The sequence } \{a_{n}\} \text{ is an arithmetic sequence.}}$$