1. Consider the face areas of the tetrahedron:
We are given that the face areas of the tetrahedron are proportional to $1, 1, \frac{1}{2}, \frac{1}{3}$. This means that if we denote the areas of the faces by $A_1, A_2, A_3, A_4$, then:
$$A_1 = k, \quad A_2 = k, \quad A_3 = \frac{k}{2}, \quad A_4 = \frac{k}{3}$$
for some positive constant $k$.

2. Verify the existence of such a tetrahedron:
To ensure that such a tetrahedron exists, we need to check that the sum of the areas of any three faces is greater than the area of the fourth face. This is a necessary condition for the face areas to form a valid tetrahedron.
$$A_1 + A_2 + A_3 = k + k + \frac{k}{2} = 2.5k > \frac{k}{3} = A_4$$
$$A_1 + A_2 + A_4 = k + k + \frac{k}{3} = 2.333k > \frac{k}{2} = A_3$$
$$A_1 + A_3 + A_4 = k + \frac{k}{2} + \frac{k}{3} = 1.833k > k = A_2$$
$$A_2 + A_3 + A_4 = k + \frac{k}{2} + \frac{k}{3} = 1.833k > k = A_1$$
Since all these inequalities hold, such a tetrahedron exists.

3. Determine the altitudes:
The altitudes of the tetrahedron are inversely proportional to the areas of the opposite faces. Therefore, if the altitudes are denoted by $h_1, h_2, h_3, h_4$, then:
$$h_1 \propto \frac{1}{A_1} = \frac{1}{k}, \quad h_2 \propto \frac{1}{A_2} = \frac{1}{k}, \quad h_3 \propto \frac{1}{A_3} = \frac{2}{k}, \quad h_4 \propto \frac{3}{k}$$
Thus, the altitudes are proportional to:
$$h_1 : h_2 : h_3 : h_4 = 1 : 1 : 2 : 3$$

4. Check if any three altitudes can form a triangle:
For three lengths to form a triangle, the sum of any two lengths must be greater than the third length. Consider the altitudes $1, 1, 2, 3$:
$$1 + 1 = 2 \not> 2$$
$$1 + 2 = 3 \not> 3$$
Therefore, no three of these altitudes can form a triangle.

Conclusion:
Since we have found a tetrahedron with altitudes proportional to $1, 1, 2, 3$ and no three of these can form a triangle, the statement is false.