Correct solutions were submitted by Paul Botham, Nancy Schwarzkopf, Lou Cairoli, Steve Young, Jens Voß.

Let's suppose that 

a = 1 + 1/2 + 1/3 + ··· + 1/k    (*)

is an integer.  Let 2^m be the largest power of 2 occurring among the denominators, i.e., 1, 2, 3, 4,..., 2^m,..., k,..., 2^m+1. The key is to observe that the least common denominator of the fractions on the right-hand side of (*) is of the form u·2^m, where u is an odd number.  Multiply both sides of (*) by u·2^m and you get an equation with integers.  On the left-hand side, (u·2^m)a is even.  On the right-hand side, each summand is even except for (u·2^m)(1/2^m) which is odd, so the right-hand side is odd.  This is, of course, impossible.

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