Non-negative integers $a, b, c, d$ satisfy the equation $a + b + c + d = 100$ and there
 exists a non-negative integer n such that
$$a+ n =b- n= c \times n = \frac{d}{n} $$
Find all 5-tuples $(a, b, c, d, n)$ satisfying all the conditions above.
1. **Analyze the second set of equations:**
   \[
   a + n = b - n = cn = \frac{d}{n}
   \]
   From this, we can infer that \(n \neq 0\) because \(n\) appears in the denominator.

2. **Express \(a, b, d\) in terms of \(n\) and \(c\):**
   Let \(a + n = k\). Then:
   \[
   a = k - n, \quad b = k + n, \quad cn = k, \quad d = kn
   \]
   Since \(cn = k\), we have \(c = \frac{k}{n}\). Also, since \(d = kn\), we have \(d = k \cdot n\).

3. **Substitute \(a, b, d\) into the first equation:**
   \[
   (k - n) + (k + n) + c + kn = 100
   \]
   Simplifying, we get:
   \[
   2k + c + kn = 100
   \]
   Since \(c = \frac{k}{n}\), substitute \(c\) into the equation:
   \[
   2k + \frac{k}{n} + kn = 100
   \]
   Multiply through by \(n\) to clear the fraction:
   \[
   2kn + k + k n^2 = 100n
   \]
   Factor out \(k\):
   \[
   k(2n + 1 + n^2) = 100n
   \]
   Solving for \(k\):
   \[
   k = \frac{100n}{2n + 1 + n^2}
   \]

4. **Determine possible values of \(n\):**
   Since \(k\) must be an integer, \(2n + 1 + n^2\) must divide \(100n\). We need to find \(n\) such that this condition holds.

   Let's test possible values of \(n\) that make \(2n + 1 + n^2\) a divisor of \(100n\):

   - For \(n = 1\):
     \[
     2(1) + 1 + 1^2 = 4 \quad \text{and} \quad k = \frac{100 \cdot 1}{4} = 25
     \]
     Thus, \(a = 24\), \(b = 26\), \(c = 25\), \(d = 25\), \(n = 1\).

   - For \(n = 4\):
     \[
     2(4) + 1 + 4^2 = 25 \quad \text{and} \quad k = \frac{100 \cdot 4}{25} = 16
     \]
     Thus, \(a = 12\), \(b = 20\), \(c = 4\), \(d = 64\), \(n = 4\).

   - For \(n = 9\):
     \[
     2(9) + 1 + 9^2 = 100 \quad \text{and} \quad k = \frac{100 \cdot 9}{100} = 9
     \]
     Thus, \(a = 0\), \(b = 18\), \(c = 1\), \(d = 81\), \(n = 9\).

### Conclusion:
The solutions are:
\[
(a, b, c, d, n) = (24, 26, 25, 25, 1), (12, 20, 4, 64, 4), (0, 18, 1, 81, 9)
\]

The final answer is \( \boxed{ (24, 26, 25, 25, 1), (12, 20, 4, 64, 4), (0, 18, 1, 81, 9) } \).