Saudi Booklet

Suppose that $k$ is the maximum index such that $a_k \le 0$ then $a_k \le 0 < a_{k+1}$. Thus $a_{k+1} + \dots + a_{50} = -(a_1 + a_2 + \dots + a_k)$ and then
$$|a_1| + |a_2| + \dots + |a_k| = |a_1 + a_2 + \dots + a_k| = a_{k+1} + a_{k+2} + \dots + a_{50} = \frac{624}{2} = 312.$$
We have
$$-312 = a_1 + a_2 + \dots + a_k \ge k \cdot a_1 \implies a_1 \le \frac{-312}{k},$$
$$312 = a_{k+1} + a_{k+2} + \dots + a_{50} \le (50-k) \cdot a_{50} \implies a_{50} \ge \frac{312}{50-k}.$$
Hence,
$$a_{50} - a_1 \ge \frac{312}{50-k} + \frac{312}{k} = 312 \left( \frac{1}{k} + \frac{1}{50-k} \right) \ge 312 \cdot \frac{4}{k+50-k} = \frac{624}{25}.$$
The minimum value of $S = \frac{624}{25}$, the equality case when
$$a_1 = a_2 = \dots = a_{25} = -\frac{312}{25} \text{ and } a_{26} = a_{27} = \dots = a_{50} = \frac{312}{25}.$$
We will continue with $T$. If there is some $a_k = 0$ then $T = 0$. Now suppose that all of given numbers are non-zero and there are $k$ negative numbers among $a_1, a_2, \dots, a_{50}$. If $k$ is odd then $T < 0$. Thus to find the maximum value of $T$, we consider $k$ is even. Put $h = 50 - k$. By AM-GM inequality, we have
$$\begin{aligned} T &\le \left( \frac{a_1 + a_2 + \dots + a_k}{k} \right)^k \cdot \left( \frac{a_{k+1} + a_{k+2} + \dots + a_{50}}{h} \right)^h \\ &= \left( \frac{312}{k} \right)^k \left( \frac{312}{h} \right)^h = \frac{312^{h+k}}{k^k h^h} = \frac{312^{50}}{k^k h^h}. \end{aligned}$$
We need to prove that
$$k^k \cdot h^h \ge 24^{24} \cdot 26^{26}.$$
WLOG, $k \le h$ then $k \le h - 2$ since $k, h$ are even. Suppose that $k \le h - 4$ then we can use local modification to prove that
$$k^k \cdot h^h > (k+1)^{k+1} \cdot (h-1)^{h-1} \iff h \cdot \left(1 + \frac{1}{h-1}\right)^{h-1} > (k+1) \cdot \left(1 + \frac{1}{k}\right)^k \quad (*)$$
Consider the function $f(x) = (x+1) \cdot (1+\frac{1}{x})^x$ with $x > 1$ and
$$g(x) = \ln f(x) = \ln(x+1) + x \cdot \ln\left(1+\frac{1}{x}\right).$$
We have
$$g'(x) = \frac{1}{1+x} + \ln\left(1 + \frac{1}{x}\right) - \frac{1}{1+x} = \ln\left(1 + \frac{1}{x}\right) \geq 0.$$
Thus $g(x)$ increasing and then (*) is proved. Therefore, the maximum value of $T$ is $13^{24} \cdot 12^{26}$. The equality case when
$$a_1 = a_2 = \dots = a_{24} = -13,\ a_{25} = a_{26} = \dots = a_{50} = 12.$$