Given $n$ positive real numbers $x_1,x_2,x_3,...,x_n$ such that
$$\left (1+\frac{1}{x_1}\right )\left(1+\frac{1}{x_2}\right)...\left(1+\frac{1}{x_n}\right)=(n+1)^n$$
Determine the minimum value of $x_1+x_2+x_3+...+x_n$.

[i]Proposed by Loh Kwong Weng[/i]
1. **Define the function and apply Jensen's Inequality:**
   Let \( f(x) = \ln\left(1 + \frac{1}{x}\right) \). The function \( f(x) \) is convex because its second derivative is positive for \( x > 0 \).

2. **Calculate the second derivative:**
   \[
   f'(x) = -\frac{1}{x(x+1)}
   \]
   \[
   f''(x) = \frac{1 + 2x}{x^2(x+1)^2}
   \]
   Since \( f''(x) > 0 \) for \( x > 0 \), \( f(x) \) is indeed convex.

3. **Apply Jensen's Inequality:**
   By Jensen's Inequality for a convex function \( f \),
   \[
   f\left(\frac{x_1 + x_2 + \cdots + x_n}{n}\right) \leq \frac{f(x_1) + f(x_2) + \cdots + f(x_n)}{n}
   \]
   Given that
   \[
   \prod_{i=1}^n \left(1 + \frac{1}{x_i}\right) = (n+1)^n
   \]
   Taking the natural logarithm on both sides, we get
   \[
   \sum_{i=1}^n \ln\left(1 + \frac{1}{x_i}\right) = n \ln(n+1)
   \]

4. **Simplify using the convexity result:**
   Using the convexity result,
   \[
   \frac{1}{n} \sum_{i=1}^n \ln\left(1 + \frac{1}{x_i}\right) = \ln(n+1)
   \]
   Therefore,
   \[
   \ln\left(1 + \frac{1}{\frac{x_1 + x_2 + \cdots + x_n}{n}}\right) \leq \ln(n+1)
   \]
   Exponentiating both sides, we get
   \[
   1 + \frac{1}{\frac{x_1 + x_2 + \cdots + x_n}{n}} \leq n+1
   \]
   Simplifying,
   \[
   \frac{1}{\frac{x_1 + x_2 + \cdots + x_n}{n}} \leq n
   \]
   \[
   \frac{n}{x_1 + x_2 + \cdots + x_n} \leq n
   \]
   \[
   x_1 + x_2 + \cdots + x_n \geq n
   \]

5. **Check if the minimum value can be achieved:**
   If \( x_1 = x_2 = \cdots = x_n = 1 \), then
   \[
   \left(1 + \frac{1}{1}\right)^n = 2^n = (n+1)^n
   \]
   This is true only for \( n = 1 \). For \( n > 1 \), we need to find \( x_i \) such that the product condition holds. However, the minimum sum \( x_1 + x_2 + \cdots + x_n \) is achieved when all \( x_i \) are equal, and each \( x_i = 1 \).

Conclusion:
\[
\boxed{n}
\]