Problem 1626

1. First, we need to determine the range of $z$ values for which the two spheres intersect. The first sphere is centered at $(0, 0, \frac{21}{2})$ with radius 6, and the second sphere is centered at $(0, 0, 1)$ with radius $\frac{9}{2}$.

2. The equation of the first sphere is:
$$x^2 + y^2 + \left(z - \frac{21}{2}\right)^2 \leq 6^2$$
The equation of the second sphere is:
$$x^2 + y^2 + (z - 1)^2 \leq \left(\frac{9}{2}\right)^2$$

3. To find the range of $z$ values, we need to consider the vertical distance between the centers of the spheres. The distance between the centers is:
$$\left| \frac{21}{2} - 1 \right| = \frac{19}{2}$$

4. The sum of the radii of the two spheres is:
$$6 + \frac{9}{2} = \frac{21}{2}$$

5. Since the distance between the centers is less than the sum of the radii, the spheres intersect. The intersection occurs within the range of $z$ values where both spheres' equations are satisfied.

6. We need to find the range of $z$ values for which the intersection occurs. The intersection will be within the range:
$$\frac{21}{2} - 6 \leq z \leq \frac{21}{2} + 6$$
Simplifying, we get:
$$\frac{9}{2} \leq z \leq \frac{33}{2}$$

7. Similarly, for the second sphere:
$$1 - \frac{9}{2} \leq z \leq 1 + \frac{9}{2}$$
Simplifying, we get:
$$-\frac{7}{2} \leq z \leq \frac{11}{2}$$

8. The intersection of these ranges is:
$$\frac{9}{2} \leq z \leq \frac{11}{2}$$
Since $z$ must be an integer, the only possible value is $z = 5$.

9. Substituting $z = 5$ into the equations of the spheres, we get:
$$x^2 + y^2 + \left(5 - \frac{21}{2}\right)^2 \leq 6^2$$
Simplifying, we get:
$$x^2 + y^2 + \left(-\frac{11}{2}\right)^2 \leq 36$$
$$x^2 + y^2 + \frac{121}{4} \leq 36$$
$$x^2 + y^2 \leq 36 - \frac{121}{4}$$
$$x^2 + y^2 \leq \frac{144}{4} - \frac{121}{4}$$
$$x^2 + y^2 \leq \frac{23}{4}$$

10. For the second sphere:
$$x^2 + y^2 + (5 - 1)^2 \leq \left(\frac{9}{2}\right)^2$$
Simplifying, we get:
$$x^2 + y^2 + 4^2 \leq \frac{81}{4}$$
$$x^2 + y^2 + 16 \leq \frac{81}{4}$$
$$x^2 + y^2 \leq \frac{81}{4} - 16$$
$$x^2 + y^2 \leq \frac{81}{4} - \frac{64}{4}$$
$$x^2 + y^2 \leq \frac{17}{4}$$

11. Notice that all $(x, y)$ satisfying the second inequality also satisfy the first one. Therefore, we need to find all the lattice points that satisfy $x^2 + y^2 \leq \frac{17}{4}$.

12. The possible integer solutions for $x$ and $y$ are:
$$(-2, 0), (2, 0), (0, -2), (0, 2), (-1, -1), (1, -1), (-1, 1), (1, 1), (-1, 0), (1, 0), (0, -1), (0, 1), (0, 0)$$

13. Each of these pairs corresponds to a point $(x, y, z)$ with $z = 5$.

Conclusion:
There are 13 lattice points in the intersection of the two spheres.

The final answer is $\boxed{13}$