Solution:

Let $d$ be the leftmost digit and $n$ the integer that remains after deleting it, so the original integer is $d \cdot 10^p + n$ for some positive integer $p.$ The condition says $d \cdot 10^p + n = 29n,$ so $d \cdot 10^p = 28n.$

Since $7 \mid 28n$ but $7 \nmid 10^p,$ the digit $d$ must be a multiple of $7,$ so $d = 7.$ Then $10^p = 4n,$ giving $n = 25 \cdot 10^{p-2},$ which requires $p \ge 2.$ The smallest case is $p = 2,$ $n = 25.$

The least such integer is $725,$ and indeed $725 = 29 \cdot 25.$