Four consecutive odd numbers, when their sum is added to their product, as well as all possible two-factor and three-factor products formed from these numbers without repetition, result in 26,879. Which four numbers are these?
Given four numbers, $a, b, c, d$, the expression in the problem, when one is added to the sum, can be transformed into a product, as follows:

$$
\begin{gathered}
1+(a+b+c+d)+(a b+a c+a d+b c+b d+c d)+ \\
+(a b c+a b d+a c d+b c d)+a b c d=(1+a)(1+b)(1+c)(1+d)
\end{gathered}
$$

which is easily observable.

If the numbers are consecutive odd numbers, then the numbers increased by one are consecutive even numbers. Denoting the middle odd number (the third in the increasing order of our odd numbers) by $x$, the even numbers are $x-3, x-1, x+1, x+3$. We know that their product is one greater than the given sum, that is,

$$
\begin{gathered}
(x-3)(x-1)(x+1)(x+3)=\left(x^{2}-9\right)\left(x^{2}-1\right)= \\
=x^{4}-10 x^{2}+9=26880
\end{gathered}
$$

From this,

$$
x^{4}-10 x^{2}-26871=0, \quad x^{2}=5 \pm 164
$$

Only the positive root, 169, corresponds to a real value of $x$, namely $x_{1}=13, x_{2}=-13$. These are odd integers, so there are two sets of four numbers that satisfy the conditions of the problem: $9,11,13,15$ and $-17,-15,-13,-11$.

Remarks. 1. In the solution, we did not use the fact that the numbers are odd integers, so it would have been sufficient to require only that they form an arithmetic sequence with a difference of 2.

We can proceed further from (11) by using these conditions and the inequality of the arithmetic and geometric means. For the positive solutions of (11), the left side is the fourth power of the geometric mean of the four consecutive even numbers. The arithmetic mean of the numbers is the odd number $x$, for which

$$
x>\sqrt[4]{26880}>\sqrt{163}>12
$$

Trying $x=13$, we get $10 \cdot 12 \cdot 14 \cdot 16=26880$, so we have a solution, $9,11,13,15$ satisfies the problem's requirements. Replacing each factor with its negative, we see that $-12-1=-13$ also provides a solution, so the numbers $-17,-15,-13,-11$ also meet the problem's requirements. Since the value of the left side of (11) is 0 for $x= \pm 1, \pm 3$, and for $x$ with an absolute value greater than 3, the absolute value of each factor, and thus the product, increases with the increase in the absolute value of $x$, there are no more solutions.