Let $p> 3$ be a prime number. Determine the number of all ordered sixes $(a, b, c, d, e, f)$ of positive integers whose sum is $3p$ and all fractions $\frac{a + b}{c + d},\frac{b + c}{d + e},\frac{c + d}{e + f},\frac{d + e}{f + a},\frac{e + f}{a + b}$ have integer values.
1. Let \( p > 3 \) be a prime number. We need to determine the number of all ordered sixes \((a, b, c, d, e, f)\) of positive integers whose sum is \(3p\) and all fractions \(\frac{a + b}{c + d}, \frac{b + c}{d + e}, \frac{c + d}{e + f}, \frac{d + e}{f + a}, \frac{e + f}{a + b}\) have integer values.

2. Let \(\frac{a + b}{c + d} = x \in \mathbb{N}\), \(\frac{c + d}{e + f} = y \in \mathbb{N}\), \(\frac{e + f}{a + b} = z \in \mathbb{N}\). Then:
   \[
   xyz = \frac{a + b}{c + d} \cdot \frac{c + d}{e + f} \cdot \frac{e + f}{a + b} = 1 \implies x = y = z = 1
   \]
   This implies that \(a + b = c + d = e + f = p\).

3. Let \(\frac{b + c}{d + e} = m \in \mathbb{N}\), \(\frac{d + e}{f + a} = n \in \mathbb{N}\). Then:
   \[
   mn + n + 1 = \frac{b + c}{d + e} \cdot \frac{d + e}{f + a} + \frac{d + e}{f + a} + 1 \implies mn + n + 1 = \frac{3p}{f + a}
   \]
   Since \(f + a \geq 2\) and \(mn + n + 1 > 1\), we have two cases:

4. **Case 1: \(f + a = p\)**
   - Then \(mn + n + 1 = 3\). Since \(m, n\) are positive integers, we must have \(m = n = 1\).
   - Hence \(b + c = d + e = f + a = p\), which gives \(a + b = b + c\), \(e + f = b + a\), \(b + c = c + d\), \(d + e = e + f\).
   - So \(a = c = e\) and \(b = d = f\).
   - All 6-tuples of the form \((a, b, c, d, e, f) = (k, p - k, k, p - k, k, p - k)\) where \(k\) is a positive integer with \(k < p\) satisfy the given conditions.
   - In conclusion, from this case, we have \(p - 1\) such 6-tuples.

5. **Case 2: \(f + a = 3\)**
   - If \(f = 1\) and \(a = 2\), we have \(e = p - 1\) and \(b = p - 2\). Additionally:
     \[
     d + e \mid b + c \implies d + p - 1 \mid p - 2 + c - d - (p - 1) \implies d + p - 1 \mid c - d - 1
     \]
     - If \(c - d - 1 \neq 0\), we must have \(d + p - 1 \leq |c - d - 1|\). But \(d + p - 1 \geq p\) and \(|c - d - 1| = |p - d - d - 1| = |p - 2d - 1| \leq p - 1\), contradiction.
     - Hence \(c - d - 1 = 0 \implies c = \frac{p + 1}{2}\) and \(d = \frac{p - 1}{2}\).
     - This subcase gives the 6-tuple \((a, b, c, d, e, f) = (2, p - 2, \frac{p + 1}{2}, \frac{p - 1}{2}, p - 1, 1)\).

   - If \(f = 2\) and \(a = 1\), we get \(e = p - 2\) and \(b = p - 1\). Also:
     \[
     d + e \mid b + c \implies d + p - 2 \mid p - 1 + c - d - (p - 2) \implies d + p - 2 \mid c - d + 1
     \]
     - If \(c - d + 1 \neq 0\), we must have \(d + p - 2 \leq |c - d + 1|\). But \(d + p - 2 \geq p - 1\) and \(|c - d + 1| = |p - d - d + 1| = |p - 2d + 1| \leq p - 1\).
     - Hence, the equality case must hold, which gives \(d = 1\) and \(c = p - 1\).
     - If \(c - d + 1 = 0\), we have \(c = \frac{p - 1}{2}\) and \(d = \frac{p + 1}{2}\).
     - This subcase gives the 6-tuples \((a, b, c, d, e, f) = (1, p - 1, p - 1, 1, p - 2, 2)\) and \((1, p - 1, \frac{p - 1}{2}, \frac{p + 1}{2}, p - 2, 2)\).

6. All in all, the requested number is \(p - 1 + 1 + 2 = p + 2\).

The final answer is \(\boxed{p + 2}\).