s = 1 + 2 + 3 + 4 + ..... + 29 + 30
t = 31 + 32 + 33 + 34 + ..... + 59 + 60

Notice that we can rewrite the second equation as follows:
t = (30 + 1) + (30 + 2) + (30 + 3) + (30 + 4) + ......+ (30 + 29) + (30 + 30)

In other words, t = (30 + 30 + 30 + 30 + 30 ... + 30 + 30) + (1 + 2 + 3 + 4 + ..... + 29 + 30)

IMPORTANT: To determine the number of 30's in the above sum, we need to use the following useful formula:
The number of integers from x to y inclusive equals y - x + 1

So, the number of integers from 31 to 60 inclusive equals 60 - 31 + 1 = 30
This means there are thirty 30's in the above sum.

That is: t = (30)(30) + (1 + 2 + 3 + 4 + ..... + 29 + 30)

So, t - s = [(30)(30) + (1 + 2 + 3 + 4 + ..... + 29 + 30)] - [1 + 2 + 3 + 4 + ..... + 29 + 30]
= (30)(30)
= 900

Answer: 900

Cheers,
Brent

I have another way to solve this it is a little quicker than above for me:
s = (n(n+1))/2 = 30x31/2 = 465
total sum of 60 integers = 60x61/2 = 1830
t = total -s = 1830-465 = 1365
now t-s = 900