Find the sphere of maximal radius that can be placed inside every tetrahedron that has all altitudes of length greater than or equal to $1.$
1. **Define the variables and relationships:**
   Let \( V \) be the volume of the tetrahedron, and let \( A, B, C, D \) be the areas of the faces of the tetrahedron. Let \( h_A, h_B, h_C, h_D \) be the altitudes to the corresponding faces of the tetrahedron. Let \( r \) be the inradius of the tetrahedron.

2. **Express the volume in terms of the altitudes and face areas:**
   The volume \( V \) of the tetrahedron can be expressed using any face area and its corresponding altitude:
   \[
   V = \frac{1}{3} A h_A = \frac{1}{3} B h_B = \frac{1}{3} C h_C = \frac{1}{3} D h_D
   \]

3. **Sum the expressions for the volume:**
   Summing these expressions, we get:
   \[
   4V = \frac{1}{3} (A h_A + B h_B + C h_C + D h_D)
   \]
   Since each altitude is at least 1, we have:
   \[
   A h_A + B h_B + C h_C + D h_D \ge A + B + C + D
   \]
   Therefore:
   \[
   4V \ge \frac{1}{3} (A + B + C + D)
   \]
   Simplifying, we get:
   \[
   V \ge \frac{A + B + C + D}{12}
   \]

4. **Express the volume in terms of the inradius:**
   The volume \( V \) can also be expressed in terms of the inradius \( r \) and the sum of the face areas:
   \[
   V = \frac{1}{3} (A + B + C + D) r
   \]

5. **Combine the inequalities:**
   Using the two expressions for \( V \), we get:
   \[
   \frac{1}{3} (A + B + C + D) r \ge \frac{A + B + C + D}{12}
   \]
   Dividing both sides by \( \frac{1}{3} (A + B + C + D) \), we obtain:
   \[
   r \ge \frac{1}{4}
   \]

6. **Verify the maximal radius:**
   We need to verify that a sphere of radius \( \frac{1}{4} \) is the largest possible sphere that can be placed in a regular tetrahedron with altitudes of length 1. For a regular tetrahedron with side length \( s \), the altitude \( h \) is given by:
   \[
   h = \frac{\sqrt{6}}{3} s
   \]
   Setting \( h = 1 \), we solve for \( s \):
   \[
   s = \frac{3}{\sqrt{6}} = \frac{\sqrt{6}}{2}
   \]
   The inradius \( r \) of a regular tetrahedron with side length \( s \) is given by:
   \[
   r = \frac{s \sqrt{6}}{12}
   \]
   Substituting \( s = \frac{\sqrt{6}}{2} \):
   \[
   r = \frac{\frac{\sqrt{6}}{2} \sqrt{6}}{12} = \frac{6}{24} = \frac{1}{4}
   \]
   Thus, the largest possible sphere that can be placed inside such a tetrahedron has radius \( \frac{1}{4} \).

The final answer is \( \boxed{ \frac{1}{4} } \)