jmc

Note that $$a_k=\frac{1}{k^2+k}=\frac{1}{k}-\frac{1}{k+1}$$for each $k.$ Thus, the sum telescopes: $$\begin{array}{rl}{a}_m+a_{m+1}+\cdots +a_{n-1}& =\left(\frac{1}{m}-\frac{1}{m+1}\right)+\left(\frac{1}{m+1}-\frac{1}{m+2}\right)+\cdots +\left(\frac{1}{n-1}-\frac{1}{n}\right)\\ & =\frac{1}{m}-\frac{1}{n}.\end{array}$$Therefore, we have the equation $1/m-1/n=1/29.$ Multiplying by $29mn$ on both sides, we have $29n-29m=mn,$ or $mn+29m-29n=0.$ Subtracting $29^2$ from both sides, we get $$(m-29)(n+29)=-29^2.$$Since $29$ is prime and $0<m<n,$ the only possibility is $m-29=-1$ and $n+29=841,$ which gives $m=28$ and $n=812.$ Thus, $m+n=28+812=\boxed{840}.$