Estonian Mathematical Olympiad

Let $e$ be the number of secondary school participants in the "Kangaroo" competition in the previous year. This year, the number of participants was greater by $\frac{1}{4}e$. Hence $e$ is divisible by 4. Since the total number of participants in the 11th and 12th grade this year was $3+4=7$ less than the number of participants in the 10th and 11th grade last year, but the number of participants from the 10th grade this year was twice the number of participants from the 12th grade last year, the total number of secondary school participants increased by a number that is 7 less than the number of participants from the 12th grade last year. Hence the number of participants from 12th grade last year was 7 more than the number of additional participants this year, or $\frac{1}{4}e+7$. The number of participants from the 10th grade this year was twice that, or $\frac{1}{2}e+14$.

The average amount of participants per grade level last year is $\frac{1}{3}e$. Since $e$ is divisible by 4, the number $\frac{1}{3}e$ can be represented as a common fraction with a numerator that is a multiple of 4 and the denominator 3. Considering all such common fractions between 60 and 65, we get three options $\frac{184}{3}$, $\frac{188}{3}$ and $\frac{192}{3}$. So $e$ can be either 184, 188 or 192 and $\frac{1}{2}e+14$ is either 106, 108 or 110 respectively.