there are k elements of order k for each k=1,2,...,n?
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No. In particular, there cannot be any elements of order n-1, since n-1 does not divide n(n+1)/2 unless n=2 or 3. If n=2, then the desired group would have order 3, hence is Z_3, which doesn't satisfy the requirements. If n=3, then the desired group would have order 6, hence is Z_6, which also doesn't satisfy the requirements.
(To see that n-1 doesn't divide n(n+1)/2 for larger n, note that n-1 and n are always relatively prime; so we would need n-1 to divide n+1, but the only prime factor these can share is 2, hence n-1 needs to be a power of 2. When n-1=2^m, n+1=2^m+2=2(2^(m-1)+1); if m>1, 2^(m-1)+1 is odd, and so for 2^m to divide 2^m+2 we would need m<=2. So we just need to check m=0,1,2 separately.)
(To see that n-1 doesn't divide n(n+1)/2 for larger n, note that n-1 and n are always relatively prime; so we would need n-1 to divide n+1, but the only prime factor these can share is 2, hence n-1 needs to be a power of 2. When n-1=2^m, n+1=2^m+2=2(2^(m-1)+1); if m>1, 2^(m-1)+1 is odd, and so for 2^m to divide 2^m+2 we would need m<=2. So we just need to check m=0,1,2 separately.)