Suppose your table or ship is supported by forces $\vec F_i$ applied at $N$ fixed points. In general, there are two vector equations you need to solve. In order for the ship to remain suspended, the net force on it must be zero:
$$\sum_i \vec F_i=-\vec G$$ where $\vec G$ is the weight. In addition, the net torque must be zero so that the table/ship does not rotate: $$\sum_i \vec r_i \times \vec F_i = 0.$$ where $\vec r_i$ are the positions at which the forces are applied, relative to the center of mass (which I'm assuming is the same as the center of gravity, i.e. the gravitational field is uniform). $\times$ here denotes cross product.
If all forces are in the same (vertical) direction, the equations you need to solve can be simplified somewhat. Namely, if $\vec F_i = F_i \hat z$ and $\vec G = -G\hat z$, $$\sum_i F_i=G$$ $$\sum_i x_iF_i=0 $$ $$\sum_i y_iF_i=0 $$
Note that this is a linear system of three equations.
1. For $N = 1$, there is an "unstable" solution if and only if $\vec r_1=0$, i.e. the force is applied at the center of mass. 766
The solution is unstable in the same sense as a one-legged table.
2. For $N=2$, there is an "unstable" solution if and only if the center of mass lies on the line connecting $\vec r_1$ and $\vec r_2$ (i.e. $\vec r_1$ and $\vec r_2$ are colinear with the center of mass). The solution is unstable in the same sense as a two-legged table.
3. For $N=3$, there is a unique, stable solution as long as no two of the points $\vec r_1$, $\vec r_2$ and $\vec r_3$ are colinear with the center of mass. If you further require that all forces $F_i$ be non-negative, the center of mass must lie in the interior of a triangle defined by $\vec r_1$, $\vec r_2$ and $\vec r_3$ as its vertices.
4. For $N\ge4$, except possibly in degenerate cases (where some pairs of "vertices" are colinear with the center of mass), there are infinitely many solutions (i.e. no unique solutions).
A four-legged table with each leg in contact with the ground doesn't actually need to be supported by equal forces applied at each leg. All that is required (besides the condition that the sum of the forces equal the weight) is that the forces on opposite legs be equal. So there are an infinite number of force distributions that will support the table. 151