How many $6$-tuples $(a, b, c, d, e, f)$ of natural numbers are there for which $a>b>c>d>e>f$ and $a+f=b+e=c+d=30$ are simultaneously true?
1. **Expressing the conditions in terms of variables:**
   Given the conditions \(a > b > c > d > e > f\) and \(a + f = b + e = c + d = 30\), we can express the numbers in terms of deviations from 15:
   \[
   a = 15 + x, \quad b = 15 + y, \quad c = 15 + z, \quad d = 15 - z, \quad e = 15 - y, \quad f = 15 - x
   \]
   Here, \(x, y, z\) are positive integers such that \(x > y > z\).

2. **Ensuring the natural number condition:**
   Since \(a, b, c, d, e, f\) are natural numbers, we need:
   \[
   1 \leq 15 - x < 15 - y < 15 - z \leq 14
   \]
   This simplifies to:
   \[
   1 \leq 15 - x \quad \text{and} \quad 15 - z \leq 14
   \]
   Therefore:
   \[
   1 \leq 15 - x \implies x \leq 14
   \]
   \[
   15 - z \leq 14 \implies z \geq 1
   \]

3. **Counting the valid tuples:**
   We need to count the number of ways to choose \(d, e, f\) such that \(d > e > f\) and \(1 \leq f < e < d \leq 14\). This is equivalent to choosing 3 distinct numbers from the set \(\{1, 2, \ldots, 14\}\) and arranging them in decreasing order.

4. **Using combinations:**
   The number of ways to choose 3 distinct numbers from 14 is given by the binomial coefficient:
   \[
   \binom{14}{3}
   \]
   Calculating this, we get:
   \[
   \binom{14}{3} = \frac{14 \times 13 \times 12}{3 \times 2 \times 1} = 364
   \]

5. **Conclusion:**
   Each choice of \(d, e, f\) uniquely determines \(a, b, c\) as \(a = 15 + x\), \(b = 15 + y\), \(c = 15 + z\) where \(x, y, z\) are the corresponding values ensuring \(a > b > c > d > e > f\).

The final answer is \(\boxed{364}\)