After dividing the equation by $xyzt$ we get $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}=1+\frac{3}{xyzt}$. Because of symmetry, without loss of generality, we can assume that
$$x \le y \le z \le t \quad \dots \tag{1}$$
from where it follows that $\frac{1}{x} \ge \frac{1}{y} \ge \frac{1}{z} \ge \frac{1}{t}$. We get $\frac{4}{x} \ge \frac{1}{x} + \frac{1}{y} + \frac{1}{z} + \frac{1}{t} = 1 + \frac{3}{xyzt} > 1$, from where we have $x < 4$.

Case 1. Let $x=3$. Then the equation is of the form $3yz + yzt + 3zt + 3yt = 3yzt + 3$, or, equivalently $3(yz + zt + yt) = 2yzt + 3$. After dividing this equation by $yzt$ we get
$$3\left(\frac{1}{y} + \frac{1}{z} + \frac{1}{t}\right) = 2 + \frac{3}{yzt} > 2, \frac{9}{y} > 2, \text{ from where we have } y \le 4.$$
The possible values for $y$ are 3 and 4.

a) For $y=4$ we get
$$3(4z + zt + 4t) = 8zt + 3,\ 12(z + t) = 5zt + 3,\ 12\left(\frac{1}{z} + \frac{1}{t}\right) = 5 + \frac{3}{zt} > 5, \frac{24}{z} > 5,$$
from where we have $z \le 4$. From (1) it follows that $z=4$ and the equation gets the form $12(4+t)=20t+3$, or, equivalently $8t=45$, which implies that $t$ is not a natural number.

b) For $y=3$, we get
$$3(3z + zt + 3t) = 6zt + 3,\ 3(z + t) = zt + 1,\ 3\left(\frac{1}{z} + \frac{1}{t}\right) = 1 + \frac{1}{zt} > 1, \frac{6}{z} > 1, z < 6.$$
The possible values for $z$ are 3, 4, 5.
- Let $z=3$. Then $3(3+t)=3t+1$ which is impossible.
- If $z=4$, then $3(4+t)=4t+1$, $t=11$.
- If $z=5$, then $3(5+t)=5t+1$, $t=7$.
We get that the quadriplets $(3,3,4,11)$, $(3,3,5,7)$ are solutions.

Case 2. Let $x=2$.
Then the equation is of the form
$$2yz + yzt + 2zt + 2yt = 2yzt + 3,$$
or, equivalently,
$$2(yz + zt + yt) = yzt + 3 \qquad (2).$$
Then each of the numbers $y,z,t$ is odd. After dividing this equation by $yzt$ we get $2\left(\frac{1}{y} + \frac{1}{z} + \frac{1}{t}\right) = 1 + \frac{3}{yzt} > 1$ from where we have $\frac{6}{y} > 1$, or, equivalently $y < 6$.

a) If $y=5$ then (2) is of the form $2(5z + zt + 5t) = 5zt + 3$, or, equivalently $10(z+t) = 3zt + 3$. Hence $10\left(\frac{1}{z} + \frac{1}{t}\right) = 3 + \frac{3}{zt} > 3$, therefore $\frac{1}{z} > \frac{3}{20}$, or equivalently $z \le 6$. The only possibility is $z=5$. We get $10(5+t) = 15t + 3$, or, equivalently $5t=47$ which implies that $t$ is not a natural number.

b) If $y=3$, (2) is of the form $2(3z + zt + 3t) = 3zt + 3$, or equivalently $6(z+t) = zt + 3$. Then $6\left(\frac{1}{z} + \frac{1}{t}\right) = 1 + \frac{3}{zt} > 1$, from where $\frac{12}{z} > 1$, or, equivalently $z < 12$. The possibilities for $z$ are 3, 5, 7, 9, 11.
- If $z=3$, then $6(3+t)=3t+3$, from where we have $3t=-15$, or equivalently $t=-5 \notin \mathbb{N}$.
- If $z=5$, then $6(5+t)=5t+3$, $t=-27 \notin \mathbb{N}$.
- If $z=7$, then $6(7+t)=7t+3$, $t=39$.
- If $z=9$, then $6(9+t)=9t+3$, from where we have $3t=51$, or, equivalently $t=17$.
Therefore in this case the solutions are the quadriplets $(2,3,7,39)$, $(2,3,9,17)$.

Case 3. The case remains when $x=1$. Then the equation is of the form $yz + yzt + zt + yt = yzt + 3$, or, equivalently $yz + zt + yt = 3$. From (1) we get $3yz \le 3$, or equivalently $yz \le 1$, from where $y=1$ and $z=1$. Then $1+2t=3$, or equivalently $t=1$. The quadriple $(1,1,1,1)$ is a solution.

Finally, the solutions to the initial equation are all permutations of $(3,3,4,11)$, $(3,3,5,7)$, $(2,3,7,39)$, $(2,3,9,17)$, $(1,1,1,1)$.