Let $M(n)=\{n,n+1,n+2,n+3,n+4,n+5\}$ be a set of 6 consecutive integers. Let's take all values of the form \[\frac{a}{b}+\frac{c}{d}+\frac{e}{f}\] with the set $\{a,b,c,d,e,f\}=M(n)$.
Let \[\frac{x}{u}+\frac{y}{v}+\frac{z}{w}=\frac{xvw+yuw+zuv}{uvw}\] be the greatest of all these values.

a) show: for all odd $n$ hold: $\gcd (xvw+yuw+zuv, uvw)=1$ iff $\gcd (x,u)=\gcd (y,v)=\gcd (z,w)=1$.
b) for which positive integers $n$ hold $\gcd (xvw+yuw+zuv, uvw)=1$?
### Part (a)
We need to show that for all odd \( n \), \(\gcd(xvw + yuw + zuv, uvw) = 1\) if and only if \(\gcd(x, u) = \gcd(y, v) = \gcd(z, w) = 1\).

1. **Expression for the sum:**
   Given the set \( M(n) = \{n, n+1, n+2, n+3, n+4, n+5\} \), we consider the expression:
   \[
   \frac{x}{u} + \frac{y}{v} + \frac{z}{w} = \frac{xvw + yuw + zuv}{uvw}
   \]
   where \( \{x, u, y, v, z, w\} = M(n) \).

2. **Greatest common divisor condition:**
   We need to show that:
   \[
   \gcd(xvw + yuw + zuv, uvw) = 1 \iff \gcd(x, u) = \gcd(y, v) = \gcd(z, w) = 1
   \]

3. **Sufficiency:**
   Assume \(\gcd(x, u) = \gcd(y, v) = \gcd(z, w) = 1\). Since \(x, y, z\) and \(u, v, w\) are distinct elements from \(M(n)\), and \(M(n)\) consists of consecutive integers, any common divisor of \(uvw\) and \(xvw + yuw + zuv\) must divide each term individually. However, since \(\gcd(x, u) = 1\), \(\gcd(y, v) = 1\), and \(\gcd(z, w) = 1\), no prime factor can divide both the numerator and the denominator simultaneously. Hence, \(\gcd(xvw + yuw + zuv, uvw) = 1\).

4. **Necessity:**
   Assume \(\gcd(xvw + yuw + zuv, uvw) = 1\). If \(\gcd(x, u) \neq 1\), then there exists a prime \(p\) such that \(p\) divides both \(x\) and \(u\). This would imply \(p\) divides \(xvw + yuw + zuv\) and \(uvw\), contradicting the assumption that \(\gcd(xvw + yuw + zuv, uvw) = 1\). Similar arguments hold for \(\gcd(y, v)\) and \(\gcd(z, w)\). Therefore, \(\gcd(x, u) = \gcd(y, v) = \gcd(z, w) = 1\).

Thus, we have shown the required equivalence.

### Part (b)
We need to determine for which positive integers \( n \) the condition \(\gcd(xvw + yuw + zuv, uvw) = 1\) holds.

1. **Expression for the maximum value:**
   We are given:
   \[
   t = \max \left( \frac{a}{b} + \frac{c}{d} + \frac{e}{f} \right) = \frac{n+5}{n} + \frac{n+4}{n+1} + \frac{n+3}{n+2}
   \]
   Simplifying the expression:
   \[
   t = \frac{(n+5)(n+1)(n+2) + (n+4)n(n+2) + (n+3)n(n+1)}{n(n+1)(n+2)}
   \]
   \[
   t = \frac{(n^2 + 7n + 10)n(n+1) + (n^2 + 6n + 8)n(n+2) + (n^2 + 5n + 6)n(n+1)}{n(n+1)(n+2)}
   \]
   \[
   t = \frac{3n^3 + 18n^2 + 28n + 10}{n(n+1)(n+2)}
   \]

2. **Greatest common divisor condition:**
   We need to check when:
   \[
   \gcd(3n^3 + 18n^2 + 28n + 10, n(n+1)(n+2)) = 1
   \]
   This holds if and only if:
   \[
   \gcd(n, 10) = 1, \quad \gcd(n+1, 3) = 1, \quad \gcd(n+2, 2) = 1
   \]

3. **Conditions on \( n \):**
   - \(\gcd(n, 10) = 1\) implies \( n \) is not divisible by 2 or 5.
   - \(\gcd(n+1, 3) = 1\) implies \( n+1 \) is not divisible by 3.
   - \(\gcd(n+2, 2) = 1\) implies \( n+2 \) is not divisible by 2, i.e., \( n \) is even.

Combining these conditions, \( n \) must be an odd number not divisible by 5 and such that \( n+1 \) is not divisible by 3.

The final answer is \( \boxed{ n } \) is an odd number not divisible by 5 and such that \( n+1 \) is not divisible by 3.