jmc

First, note the well-known formula that $1+2+3+\cdots +n=\frac{n(n+1)}{2}$. Thus we are looking for $n$ such that $\frac{n(n+1)}{2}$ is a perfect square. Since $n$ and $n+1$ are relatively prime, the odd one will have to be a perfect square, and the even one will have to be twice a perfect square. Thus we are looking for solutions to $a^2-2b^2=\pm 1$ where $a^2\ge 4$. It's clear $b=1$ does not work, but trying $b=2$, we find that $2(2^2)+1=9=3^2$. Therefore, $n=2(2^2)=\boxed{8}$ is the smallest solution. We may confirm our answer by checking that ${\left(\frac{n(n+1)}{2}\right)}^2=36^2$ when $n=8$, and indeed $36^2=6^4$ is a perfect fourth power.