Let $S$ be the set of all students. For $s \in S$ we denote by $a_s$ the number of students in the group of the student $s$ in the first distribution and by $b_s$ the number of students in the group of student $s$ in the second distribution.
We have
$$\sum_{s \in S} \frac{1}{a_s} = N, \quad \sum_{s \in S} \frac{1}{b_s} = N + K.$$
Hence the sum of differences of numbers assigned to each student is equal to
$$\sum_{s \in S} \left( \frac{1}{b_s} - \frac{1}{a_s} \right) = K.$$
Since $\left| \frac{1}{b_s} - \frac{1}{a_s} \right| < 1$ it follows that for at least $K + 1$ students we have $\frac{1}{b_s} - \frac{1}{a_s} > 0$, i.e. $b_s < a_s$.