Solution: Method 1: First, arrange four students to take morning tests on "height and weight", "standing long jump", "vital capacity" and "steps". There are 44 different arrangements. Next, in the afternoon, we will arrange tests on "height and weight", "standing long jump", "vital capacity" and "grip strength". Suppose students A, B and C arrange the tests of "height and weight", "standing long jump" and "vital capacity" in the morning respectively. If student D chooses the "grip strength" test, students A, B and C are arranged to conduct cross tests respectively. If student D chooses 1 in the tests of "height and weight", "standing long jump" and "vital capacity", there are 3 1 ways, and there are 3 ways to arrange students A, B and C for testing. According to the counting principle, the number of species arranged in * * * is A44(2+A3 1×3)=264.

So the answer is 264.

Solution: Suppose there is no such restriction: the "grip strength" item is not measured in the morning and the "step" item is not measured in the afternoon. You can choose five items in the morning and only four items in the afternoon. The number of test items in the morning is 4×5=20, and the number of test items in the afternoon is 4×4= 16, so we can easily get the combination.

Consider this restriction again: the "grip strength" item is not measured in the morning and the "step" item is not measured in the afternoon. Among 320 combinations, how many types of grip strength are there in the morning? Very easy to calculate, the total is110,32; What are the combination of steps in the afternoon, which is also 1 10, 32 of the total. So 320-32-32=256 species.

But in the end, we still have to consider 64 kinds of repeated removal, such as the combination of a classmate's grip strength in the morning and the number of steps in the afternoon (this was removed twice) and the combination of a classmate's grip strength in the morning and afternoon (also removed twice). B.C.D is needed in this case, so 2× 4 = 8 needs to be added.

So the final calculation result is 4× 5× 4× 4-32-32+8 = 264.

Answer: 264.

This paper examines permutation and combination and its application, focusing on the application of reasoning analysis and classified discussion thinking.