For each positive integer $n$, find the largest real number $C_n$ with the following property. Given any $n$ real-valued functions $f_1(x), f_2(x), \cdots, f_n(x)$ defined on the closed interval $0 \le x \le 1$, one can find numbers $x_1, x_2, \cdots x_n$, such that $0 \le x_i \le 1$ satisfying 
\[|f_1(x_1)+f_2(x_2)+\cdots f_n(x_n)-x_1x_2\cdots x_n| \ge C_n\]

[i]Marko Radovanović, Serbia[/i]
To find the largest real number \( C_n \) such that for any \( n \) real-valued functions \( f_1(x), f_2(x), \ldots, f_n(x) \) defined on the closed interval \( 0 \le x \le 1 \), there exist numbers \( x_1, x_2, \ldots, x_n \) such that \( 0 \le x_i \le 1 \) and 
\[ \left| f_1(x_1) + f_2(x_2) + \cdots + f_n(x_n) - x_1 x_2 \cdots x_n \right| \ge C_n, \]
we proceed as follows:

1. **Assume the contrary**: Suppose there exists a set of functions \( f_1, f_2, \ldots, f_n \) such that for all \( x_1, x_2, \ldots, x_n \) in \([0, 1]\),
\[ \left| f_1(x_1) + f_2(x_2) + \cdots + f_n(x_n) - x_1 x_2 \cdots x_n \right| < C_n. \]

2. **Define the range of each function**: Let \( I_k \) be the length of the range of \( f_k \), i.e., \( I_k = \max(f_k) - \min(f_k) \).

3. **Consider the sum of the first \( n-1 \) functions**: Notice that
\[ \left| \sum_{k=1}^{n-1} f_k(x_k) + f_n(0) \right| < C_n. \]
The range of the function \( g(x_1, \ldots, x_{n-1}) = \sum_{k=1}^{n-1} f_k(x_k) \) has length \( I_1 + I_2 + \cdots + I_{n-1} = I - I_n \), where \( I = \sum_{k=1}^n I_k \).

4. **Inequality for the range**: The above inequality implies that the length of the range of \( g \) is less than \( 2C_n \), since \( f_n(0) \) is constant. Therefore,
\[ I - I_n < 2C_n. \]

5. **Range of individual functions**: For some \( k \), \( I_k < \frac{2C_n}{n-1} \).

6. **Consider the function \( f_k(x) \)**: Notice that
\[ \left| f_k(x) + D - x \right| < C_n, \]
where \( D = \sum_{i \neq k} f_i(1) \). This must hold for all \( x \). Therefore, the range of \( g(x) = f_k(x) - x \) has length at most \( 2C_n \).

7. **Range constraints**: Since \( I_k < \frac{2C_n}{n-1} \), assume \( a \le f_k(x) \le a + \frac{2C_n}{n-1} \) for all \( x \). Then,
\[ g(1) \le a - 1 + \frac{2C_n}{n-1} \]
and
\[ g(0) \le 1. \]
Hence,
\[ 1 - \frac{2C_n}{n-1} < 2C_n. \]

8. **Solve the inequality**: From the above inequality,
\[ 1 - \frac{2C_n}{n-1} < 2C_n \]
\[ 1 < 2C_n + \frac{2C_n}{n-1} \]
\[ 1 < 2C_n \left( 1 + \frac{1}{n-1} \right) \]
\[ 1 < 2C_n \left( \frac{n}{n-1} \right) \]
\[ \frac{n-1}{n} < 2C_n \]
\[ \frac{n-1}{2n} < C_n. \]

Thus, we have shown that \( C_n \le \frac{n-1}{2n} \).

9. **Construction of functions**: Consider the functions \( f_i(x) = \frac{x^n}{n} - D \) for all \( i \), where \( D \) is a constant. This construction ensures that the range of each \( f_i \) is appropriately scaled.

10. **Verification**: To verify that \( \left| f_1(x_1) + f_2(x_2) + \cdots + f_n(x_n) - x_1 x_2 \cdots x_n \right| \le C_n \) is an elementary inequality, we note that the choice of \( f_i(x) = \frac{x^n}{n} - D \) ensures the desired properties.

Therefore, the largest real number \( C_n \) is \( \boxed{\frac{n-1}{2n}} \).