Find all relatively distinct integers $m, n, p\in \mathbb{Z}_{\ne 0}$ such that the polynomial
$$F(x) = x(x - m)(x - n)(x - p) + 1$$is reducible in $\mathbb{Z}[x].$
To determine all relatively distinct integers \( m, n, p \in \mathbb{Z}_{\ne 0} \) such that the polynomial 
\[ F(x) = x(x - m)(x - n)(x - p) + 1 \]
is reducible in \(\mathbb{Z}[x]\), we will consider two main cases: when \( F(x) \) has a linear factor and when \( F(x) can be factored into two quadratic polynomials.

### Case 1: \( F(x) \) has a linear factor
If \( x - a \) is a factor of \( F(x) \), then \( F(a) = 0 \). This implies:
\[ F(a) = a(a - m)(a - n)(a - p) + 1 = 0 \]
\[ a(a - m)(a - n)(a - p) = -1 \]

#### Subcase 1.1: \( a = 1 \)
\[ 1(1 - m)(1 - n)(1 - p) = -1 \]
\[ (1 - m)(1 - n)(1 - p) = -1 \]

Since \( m, n, p \) are distinct integers, it is impossible for the product of three distinct integers to be \(-1\). Therefore, there are no solutions in this subcase.

#### Subcase 1.2: \( a = -1 \)
\[ (-1)(-1 - m)(-1 - n)(-1 - p) = -1 \]
\[ (m + 1)(n + 1)(p + 1) = -1 \]

Similarly, since \( m, n, p \) are distinct integers, it is impossible for the product of three distinct integers to be \(-1\). Therefore, there are no solutions in this subcase.

### Case 2: \( F(x) \) can be factored into two quadratic polynomials
Assume \( F(x) \) can be factored as:
\[ F(x) = (x^2 + ax + b)(x^2 + cx + d) \]

By comparing coefficients, we have:
\[ F(x) = x^4 + (a + c)x^3 + (ac + b + d)x^2 + (ad + bc)x + bd \]
\[ F(x) = x^4 - (m + n + p)x^3 + (mn + mp + np)x^2 - (mnp)x + 1 \]

Equating coefficients, we get:
1. \( a + c = -(m + n + p) \)
2. \( ac + b + d = mn + mp + np \)
3. \( ad + bc = -mnp \)
4. \( bd = 1 \)

Since \( g(0)h(0) = 1 \), \( g(m)h(m) = 1 \), \( g(n)h(n) = 1 \), and \( g(p)h(p) = 1 \), we have \( g(t) = h(t) = \pm 1 \) for \( t = 0, m, n, p \).

#### Subcase 2.1: \( g(0) = g(m) = 1 \) and \( g(n) = g(p) = -1 \)
Let \( g(x) = x(x - m) + 1 \) and \( h(x) = (x - n)(x - p) - 1 \).

Equating the polynomials:
\[ x(x - m) + 1 = (x - n)(x - p) - 1 \]
\[ x^2 - mx + 1 = x^2 - (n + p)x + np - 1 \]

By comparing coefficients:
\[ -m = -(n + p) \]
\[ m = n + p \]
\[ 1 = np - 1 \]
\[ np = 2 \]

Possible pairs \((n, p)\) that satisfy \( np = 2 \) are \((1, 2)\) and \((-1, -2)\).

For \((n, p) = (1, 2)\):
\[ m = 1 + 2 = 3 \]
Thus, \((m, n, p) = (3, 1, 2)\).

For \((n, p) = (-1, -2)\):
\[ m = -1 + (-2) = -3 \]
Thus, \((m, n, p) = (-3, -1, -2)\).

#### Subcase 2.2: \( g(0) = g(m) = -1 \) and \( g(n) = g(p) = 1 \)
Let \( g(x) = x(x - m) - 1 \) and \( h(x) = (x - n)(x - p) + 1 \).

Equating the polynomials:
\[ x(x - m) - 1 = (x - n)(x - p) + 1 \]
\[ x^2 - mx - 1 = x^2 - (n + p)x + np + 1 \]

By comparing coefficients:
\[ -m = -(n + p) \]
\[ m = n + p \]
\[ -1 = np + 1 \]
\[ np = -2 \]

Possible pairs \((n, p)\) that satisfy \( np = -2 \) are \((-1, 2)\) and \((1, -2)\).

For \((n, p) = (-1, 2)\):
\[ m = -1 + 2 = 1 \]
Thus, \((m, n, p) = (1, -1, 2)\).

For \((n, p) = (1, -2)\):
\[ m = 1 + (-2) = -1 \]
Thus, \((m, n, p) = (-1, 1, -2)\).

### Conclusion
The possible sets of \((m, n, p)\) are:
\[ (3, 1, 2), (-3, -1, -2), (1, -1, 2), (-1, 1, -2) \]

The final answer is \( \boxed{ (3, 1, 2), (-3, -1, -2), (1, -1, 2), (-1, 1, -2) } \) and their permutations.