International Mathematical Olympiad

The sum of the first 100 positive integers is 5050. The sum of the first n positive integers is n(n+1)/2. Therefore, we have 5050 = n(n+1)/2. Solving for n, we get n = 200.

The maximum possible value of $a_1^2 + a_2^2 + ... + a_n^2$ is achieved when $a_1 = a_2 = ... = a_n = 2023/n$. Therefore, the maximum possible value is $n(2023/n)^2 = 2023^2/n = 8092.

The area of triangle DEF is equal to the area of triangle ABC minus the areas of triangles ABD, BCE, and ACF. The area of triangle ABC is (1/2)(13)(14) = 91. The area of triangle ABD is (1/2)(5)(12) = 30. The area of triangle BCE is (1/2)(4)(9) = 18. The area of triangle ACF is (1/2)(3)(12) = 18. Therefore, the area of triangle DEF is 91 - 30 - 18 - 18 = 24.

Let $y = f(x)$. Then the equation $f(x) = f(f(x))$ becomes $y = f(y)$. This is a quadratic equation in $y$. Solving for $y$, we get $y = 1$ or $y = 2$. Therefore, there are two real solutions of the equation $f(x) = f(f(x))$.

The maximum value of $a^2 + b^2 + c^2$ is achieved when $a = b = c = 1$. Therefore, the maximum value is $1^2 + 1^2 + 1^2 = 3.

There are four ways to express $n$ as a sum of two squares: $n = a^2 + b^2$, $n = a^2 + (b^2 + c^2)$, $n = (a^2 + b^2) + c^2$, and $n = (a^2 + b^2 + c^2) + d^2$, where $a, b, c, d$ are positive integers.

The sum of all the elements of $S$ is equal to the sum of the multiples of 3 less than 100 plus the sum of the multiples of 5 less than 100 minus the sum of the multiples of 15 less than 100. The sum of the multiples of 3 less than 100 is 3 + 6 + 9 + ... + 96 + 99 = 1683. The sum of the multiples of 5 less than 100 is 5 + 10 + 15 + ... + 90 + 95 = 2223. The sum of the multiples of 15 less than 100 is 15 + 30 + 45 + ... + 75 + 90 = 360. Therefore, the sum of all the elements of $S$ is 1683 + 2223 - 360 = 1863.