Solution:

Since $\frac{1}{k^2 + k} = \frac{1}{k(k + 1)} = \frac{1}{k} - \frac{1}{k + 1},$ the sum telescopes: $$a_m + a_{m+1} + \cdots + a_{n-1} = \frac{1}{m} - \frac{1}{n} = \frac{1}{29}.$$

Multiplying through by $29mn$ gives $29n - 29m = mn,$ which rearranges to $(29 - m)(29 + n) = 29^2.$ Since $29$ is prime and $29 + n \gt 29,$ the only factorization with $m$ a positive integer is $29 - m = 1$ and $29 + n = 841,$ so $m = 28$ and $n = 812.$

Therefore $m + n = 28 + 812 = 840.$