n個連續數之和係咪比n除到?

It must be true when n is odd and it must be false when n is even.

You should know that 1+2+3+...+x=x(x+1)/2

Proof:
For any inerger a,
　(a+1)+(a+2)+(a+3)+...+(a+n)
= (a+a+a+...+a)+(1+2+3+...+n)
= an+n(n+1)/2
= n[a+(n+1)/2]

If n is odd, n+1 must be even, i.e. divisible by 2.
Hence (n+1)/2 must be an integer.
This implies that a+(n+1)/2 is also an integer.(Note that a is an integer)
So it must be true when n is odd.

However, when n is even, n+1 must be odd, i.e. not divisible by 2.
Hence (n+1)/2 can never be an integer.
This implies that a+(n+1)/2 is also NOT an integer.(Note that a is an integer)
So it must be false when n is even.

If n is odd, it is true, if n is even, it is not true. First of all, please understand the following result:
1 + 2 + 3 + 4 + 5 + .......+ m = m(m+1)/2. When divided by m, the answer is
[m(m+1)/2]/m = m(m+1)/2m = (m+1)/2. This is an integer if m is odd.
Now for
(n+1) + (n+2) + (n+3) + (n+4) + ...... to m number of terms.
= nm + 1 + 2 + 3 + 4 + 5 + .......+ m
= nm + m(m+1)/2. When divided by m
= n + (m+1)/2.
So if m is odd, n+(m+1)/2 is an integer, that is divisible by m.