Can we express all 4-digit natural number with sums of 3 squared number? Why or why not? thanks?

No it is not true. There are many such numbers. I can't give a thorough explanation for why not, but starting with 0, and making these sums, it becomes apparent that there are many such numbers. The gaps are just too high and they spread out even more since n^2 increases at a rate of 2n.

The first such number is 1007.

At first glance you see 100 C3 = ~160000 which should be plenty to cover 10000 numbers.

But then looking at mod 4, you see that n=3 mod 4 must be the sum of 3 odd squares, while n=0 mod 4 must be the sum of three even squares, and n=2 mod 4 is the sum of two odds and an even, and n=1 mod 4 is the sum of an odd and two evens. Then for each one you have 2500 numbers, and only 50C3= ~20000 possibilities. Just an 8-1 ratio, and the overlaps and out of bounds considerations are enough to create gaps.

I would like to know the answer to this great question.