Amanita, let's carefully solve the equation inside the green rectangle:

$\left(5 + \frac{3}{m}\right)\left(n + \frac{1}{2}\right) = 19$

Here’s my step-by-step plan:

Step 1: Expand the left side of the equation using the distributive property.
Step 2: Rearrange the equation to isolate one variable in terms of the other.
Step 3: Check for integer solutions for $m$ and $n$, since the equation looks like it might expect whole numbers.
Step 4: Substitute possible values and verify which ones satisfy the equation.

Let's start with Step 1: Expand the left side.

$\left(5 + \frac{3}{m}\right)\left(n + \frac{1}{2}\right) = 19$

Expanding gives:

$5\left(n + \frac{1}{2}\right) + \frac{3}{m}\left(n + \frac{1}{2}\right) = 19$

$5n + \frac{5}{2} + \frac{3n}{m} + \frac{3}{2m} = 19$

Now, let's move to Step 2: Rearranging the equation to group like terms and isolate variables.

$5n + \frac{5}{2} + \frac{3n}{m} + \frac{3}{2m} = 19$

Let's subtract $\frac{5}{2}$ from both sides:

$5n + \frac{3n}{m} + \frac{3}{2m} = 19 - \frac{5}{2}$

Now, let's calculate $19 - \frac{5}{2}$:

$$19 - \frac{5}{2}$$

I will calculate this value. Amanita, after subtracting, we have:

$5n + \frac{3n}{m} + \frac{3}{2m} = \frac{33}{2}$

Now, let's move to Step 3: Combine the terms with $n$ and group the fractions for easier substitution.

$5n + \frac{3n}{m} = \frac{33}{2} - \frac{3}{2m}$

Let's factor $n$ on the left:

$n\left(5 + \frac{3}{m}\right) = \frac{33}{2} - \frac{3}{2m}$

Now, let's solve for $n$:

$n = \frac{\frac{33}{2} - \frac{3}{2m}}{5 + \frac{3}{m}}$

Let's simplify the denominator $5 + \frac{3}{m}$:

$5 + \frac{3}{m} = \frac{5m + 3}{m}$

Now, substitute this into the equation for $n$:

$n = \frac{\frac{33}{2} - \frac{3}{2m}}{\frac{5m + 3}{m}}$

Let's simplify the numerator $\frac{33}{2} - \frac{3}{2m}$:

$\frac{33}{2} - \frac{3}{2m} = \frac{33m - 3}{2m}$

Now, substitute both simplified numerator and denominator:

$n = \frac{\frac{33m - 3}{2m}}{\frac{5m + 3}{m}}$

This simplifies to:

$n = \frac{33m - 3}{2m} \times \frac{m}{5m + 3}$

$n = \frac{33m - 3}{2(5m + 3)}$

Now, let's check for integer values of $m$ that make $n$ an integer. Let's try small positive integer values for $m$ and see if $n$ is a whole number. I will check $m = 1, 2, 3, 4, 5$.

Let's calculate $n$ for $m = 1$:

$n = \frac{33 \times 1 - 3}{2(5 \times 1 + 3)}$

$n = \frac{33 - 3}{2(5 + 3)}$

$$n = \frac{30}{16}$$

Let's check if this is an integer. Amanita, for $m = 1$, $n = \frac{15}{8}$, which is not a whole number.

Let's try $m = 2$:

$n = \frac{33 \times 2 - 3}{2(5 \times 2 + 3)}$

$n = \frac{66 - 3}{2(10 + 3)}$

$$n = \frac{63}{26}$$

Let's check if this is an integer. Amanita, for $m = 2$, $n = \frac{63}{26}$, which is also not a whole number.

Let's try $m = 3$:

$n = \frac{33 \times 3 - 3}{2(5 \times 3 + 3)}$

$n = \frac{99 - 3}{2(15 + 3)}$

$$n = \frac{96}{36}$$

Let's check if this is an integer. Amanita, for $m = 3$, $n = \frac{8}{3}$, which is not a whole number.

Let's try $m = 4$:

$n = \frac{33 \times 4 - 3}{2(5 \times 4 + 3)}$

$n = \frac{132 - 3}{2(20 + 3)}$

$$n = \frac{129}{46}$$

Let's check if this is an integer. Amanita, for $m = 4$, $n = \frac{129}{46}$, which is not a whole number.

Let's try $m = 5$:

$n = \frac{33 \times 5 - 3}{2(5 \times 5 + 3)}$

$n = \frac{165 - 3}{2(25 + 3)}$

$$n = \frac{162}{56}$$

Let's check if this is an integer. Amanita, for $m = 5$, $n = \frac{81}{28}$, which is not a whole number.

Now, let's try a negative value, $m = -1$:

$n = \frac{33 \times (-1) - 3}{2(5 \times (-1) + 3)}$

$n = \frac{-33 - 3}{2(-5 + 3)}$

$$n = \frac{-36}{2 \times -2}$$ $$n = \frac{-36}{-4}$$

Let's check if this is an integer. Amanita, for $m = -1$, $n = 9$, which is a whole number!

So, the solution is:

$$m = -1, \quad n = 9$$

This pair of values satisfies the original equation.