100

Another way to do it: 1+99=100, 2+98=100...+,49+51 , which is 4900 and then add the remaining 50.

Using formula for the sum of first $n$ integers, which is $\dfrac{n(n+1)}2$

$1+2+3+...+99=\dfrac{99(99+1)}2=\boxed{4950}$

Relevant wiki: Arithmetic Progression

It is an Arithmetic Progression with $d=1, n=99,a_1=1$ and $a_8=99$ . We want to find the sum of terms, so

$s_n=\dfrac{n}{2}(a_1+a_n)$ $\implies$ $s_8=\dfrac{n}{2}(a_1+a_8)=\dfrac{99}{2}(1+99)=\color{#D61F06}\boxed{4950}$