11.4. Instead of crosses in the expression $x \cdot x + x \cdot x + \ldots + x \cdot x$ (50 terms), the numbers $1, \ldots, 100$ were placed, each exactly once. What is the maximum and minimum value that the resulting expression can have?

ANSWER: $169150, 85850$.

CRITERIA

The score was assigned based on the sum of advancements in three directions, constituting a complete solution: guessing the required arrangements of numbers, justifying the maximality and minimality of the resulting expressions, and calculating these expressions.

Guessing: 0 - no more than one of the arrangements was guessed, 1 - both were guessed.

Justification: 0 - the justification is incorrect, missing, or relies only on "transinequality" without additional reasoning.

1 - For 4 numbers $a < b < c < d$, it is proven that $a b + c d$ and $a d + b c$ are the maximum and minimum, respectively. The transition to the case of 100 numbers is either absent, replaced by the word "analogously," or incorrect.

2 - justification with a minor gap.

3 - complete justification.

Calculation: 0 - numerical answers were either not obtained or were incorrect and significantly different from the correct ones (deviation of each answer from the correct one is more than 1000).

1 - numerical answer for at least one arrangement was obtained and differs from the correct one by no more than 1000.

2 - both numerical answers were obtained, and the total deviation from the correct ones is no more than 1000.

3 - both numerical answers were obtained correctly without errors.

Scoring based on the sum of advancements in the three specified directions:

| 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| ---: | ---: | ---: | ---: | ---: | ---: | ---: | ---: |
| - | $-/$. | $-/+$ | $+/ 2$ | $+/-$ | $+/$. | $+/$. | + |
Solution. Let $aa c+b d>a d+b c$. The first inequality is obtained by expanding the brackets in the inequality $(d-a)(c-b)>0$, the second - in the inequality $(b-a)(d-c)>0$.

Then, for every arrangement of numbers with a term of the form $a b+c d$ or $a c+b d$, we can associate a smaller value by substituting $a d+b c$ in this place. Take the arrangement with the minimum value, and consider the multiplier of the number 1. If it is not equal to 100, then the quartet of numbers from 1, 100, and their multipliers has the form $a b+c d$ or $a c+b d$, which contradicts the minimality of the value. Thus, we have the term $1 \cdot 100$. Reasoning similarly, the multiplier of 2 must be 99, and so on. We get the sum $1 \cdot 100+2 \cdot 99+\cdots+50 \cdot 51$, which is equal to

$$
\sum_{n=1}^{50} n \cdot(101-n)=101\left(\sum_{n=1}^{50} n\right)-\sum_{n=1}^{50} n^{2}
$$

Using the formulas for the sum of an arithmetic progression and the sum of consecutive squares, we get that it is equal to

$$
101 \cdot \frac{50 \cdot 51}{2}-\frac{50 \cdot 51 \cdot(2 \cdot 50+1)}{6}=85850
$$

On the other hand, for every arrangement of numbers with a term of the form $a d+b c$ or $a c+b d$, we can associate a larger value by substituting $a b+c d$ in this place. Take the arrangement with the maximum value, and consider the multiplier of the number 1. If it is not equal to 2, then the quartet of numbers from 1, 2, and their multipliers has the form $a d+b c$ or $a c+b d$, which contradicts the maximality of the value. Thus, we have the term $1 \cdot 2$. Reasoning similarly, the multiplier of 3 must be 4, and so on. We get the sum $1 \cdot 2+3 \cdot 4+\cdots+99 \cdot 100$, which is equal to

$$
\sum_{n=1}^{50}(2 n-1) \cdot 2 n=4\left(\sum_{n=1}^{50} n^{2}\right)-2\left(\sum_{n=1}^{50} n\right)=4 \cdot \frac{50 \cdot 51 \cdot(2 \cdot 50+1)}{6}-2 \cdot \frac{50 \cdot 51}{2}=169150
$$

Second Method (proposed by an olympiad participant)

Let $A=1^{2}+2^{2}+\cdots+100^{2}, B=x_{1} y_{1}+\cdots+x_{50} y_{50}$ - the arrangement from the problem's condition. Consider the difference

$$
A-2 B=\left(x_{1}-y_{1}\right)^{2}+\cdots+\left(x_{50}-y_{50}\right)^{2} .
$$

Since $A$ is a constant number, $B$ will be maximal when $A-2 B$ is minimal. Since in each bracket there is the difference of two different natural numbers, $A-2 B \geq 50$, and this value is achieved with the arrangement $x_{i}=2 i-1, y_{i}=2 i$. Therefore,

$$
A-2 B_{\max }=50 \Rightarrow B_{\max }=\frac{A-50}{2}=\frac{1}{2}\left(\frac{100 \cdot 101 \cdot 201}{6}-50\right)=169150
$$

Consider the sum:

$$
A+2 B=\left(x_{1}+y_{1}\right)^{2}+\cdots+\left(x_{50}+y_{50}\right)^{2}
$$

$B$ is minimal when $A+2 B$ is minimal. From the inequality between the quadratic mean and the arithmetic mean, it follows that

$$
A+2 B \geq 50\left(\frac{x_{1}+y_{1}+\cdots+x_{50}+y_{50}}{50}\right)^{2}=50 \cdot 101^{2}
$$

Equality is achieved when all terms $\left(x_{i}+y_{i}\right)$ are equal, i.e., when $x_{i}=i, y_{i}=101-i$. Therefore,

$$
A+2 B_{\min }=50 \cdot 101^{2} \Rightarrow B_{\min }=\frac{1}{2}\left(50 \cdot 101^{2}-\frac{100 \cdot 101 \cdot 201}{6}\right)=85850
$$