For each positive integer $n$ we consider the function $f_{n}:[0,n]\rightarrow{\mathbb{R}}$ defined by $f_{n}(x)=\arctan{\left(\left\lfloor x\right\rfloor \right)}  $, where $\left\lfloor x\right\rfloor $ denotes the floor of the real number $x$. Prove that $f_{n}$ is a Riemann Integrable function and find $\underset{n\rightarrow\infty}{\lim}\frac{1}{n}\int_{0}^{n}f_{n}(x)\mathrm{d}x.$
1. **Prove that \( f_n \) is Riemann integrable:**

   The function \( f_n(x) = \arctan(\lfloor x \rfloor) \) is defined on the interval \([0, n]\). The floor function \(\lfloor x \rfloor\) is piecewise constant with discontinuities at integer points. Therefore, \( f_n(x) \) is also piecewise constant with discontinuities at integer points \( k \) where \( k \in \{0, 1, 2, \ldots, n-1\} \).

   Since the set of discontinuities of \( f_n \) is finite (specifically, there are \( n \) discontinuities), and each discontinuity is a jump discontinuity, \( f_n \) is Riemann integrable on \([0, n]\) by the criterion that a function with a finite number of discontinuities on a closed interval is Riemann integrable.

2. **Evaluate the integral \( \int_0^n f_n(x) \, \mathrm{d}x \):**

   We can break the integral into a sum of integrals over each interval \([k, k+1)\) for \( k = 0, 1, 2, \ldots, n-1 \):
   \[
   \int_0^n f_n(x) \, \mathrm{d}x = \sum_{k=0}^{n-1} \int_k^{k+1} \arctan(k) \, \mathrm{d}x
   \]
   Since \( f_n(x) = \arctan(k) \) is constant on each interval \([k, k+1)\), we have:
   \[
   \int_k^{k+1} \arctan(k) \, \mathrm{d}x = \arctan(k) \cdot (k+1 - k) = \arctan(k)
   \]
   Therefore,
   \[
   \int_0^n f_n(x) \, \mathrm{d}x = \sum_{k=0}^{n-1} \arctan(k)
   \]

3. **Find the limit \( \lim_{n \to \infty} \frac{1}{n} \int_0^n f_n(x) \, \mathrm{d}x \):**

   We need to evaluate:
   \[
   a_n = \frac{1}{n} \int_0^n f_n(x) \, \mathrm{d}x = \frac{1}{n} \sum_{k=0}^{n-1} \arctan(k)
   \]

   To find the limit as \( n \to \infty \), we use the fact that the sum \( \sum_{k=0}^{n-1} \arctan(k) \) can be approximated by an integral. Specifically, we use the asymptotic behavior of the arctangent function. For large \( k \), \( \arctan(k) \approx \frac{\pi}{2} - \frac{1}{k} \).

   Therefore,
   \[
   \sum_{k=0}^{n-1} \arctan(k) \approx \sum_{k=0}^{n-1} \left( \frac{\pi}{2} - \frac{1}{k} \right)
   \]
   The sum of \( \frac{\pi}{2} \) terms is:
   \[
   \sum_{k=0}^{n-1} \frac{\pi}{2} = n \cdot \frac{\pi}{2}
   \]
   The sum of \( \frac{1}{k} \) terms is the harmonic series, which is approximately \( \ln(n) \):
   \[
   \sum_{k=1}^{n-1} \frac{1}{k} \approx \ln(n)
   \]
   Therefore,
   \[
   \sum_{k=0}^{n-1} \arctan(k) \approx n \cdot \frac{\pi}{2} - \ln(n)
   \]

   Dividing by \( n \) and taking the limit as \( n \to \infty \):
   \[
   a_n = \frac{1}{n} \sum_{k=0}^{n-1} \arctan(k) \approx \frac{1}{n} \left( n \cdot \frac{\pi}{2} - \ln(n) \right) = \frac{\pi}{2} - \frac{\ln(n)}{n}
   \]
   As \( n \to \infty \), \( \frac{\ln(n)}{n} \to 0 \). Therefore,
   \[
   \lim_{n \to \infty} a_n = \frac{\pi}{2}
   \]

The final answer is \( \boxed{ \frac{\pi}{2} } \)