2. Among all quadruples of natural numbers $(k, l, m, n), k>l>m>n$, find the one such that the sum $\frac{1}{k}+\frac{1}{l}+\frac{1}{m}+\frac{1}{n}$ is less than one and as close to it as possible.
Solution. First, note that using a "greedy algorithm" (i.e., sequentially choosing the numbers $n, m, l, k$ to be as large as possible), we obtain the quadruple $(43,7,3,2):$

$$
1 / 43 + 1 / 7 + 1 / 3 + 1 / 2 = 1805 / 1806
$$

Thus, for the desired quadruple $(k, l, m, n)$, we have $1 / k + 1 / l + 1 / m + 1 / n \geq 1805/1806$. The left side of this inequality is bounded above by $4/n$. Therefore, $n \leq 3$, meaning $n=2$ or $n=3$.

Suppose $n=3$. Then $3 / m \geq 1805 / 1806 - 1 / 3$, so $m \leq 4$. Thus, $m=4$. Next, $2 / l \geq 1805 / 1806 - 1 / 3 - 1 / 4$, so $l \leq 4$, which is a contradiction.

Therefore, $n=2$. Assume $m=4$. Then $2 / l \geq 1805 / 1806 - 1 / 2 - 1 / 4$, so $l \leq 8$. Additionally, $l \geq 5$ (since $l > m$) and $1 / k \geq 1805 / 1806 - 1 / 2 - 1 / 4 - 1 / 5$, so $k \leq 20$. This means $k l m n \leq 2 \cdot 4 \cdot 8 \cdot 20 = 12801 / 1806$ and the quadruple $(k, l, m, n)$ is not optimal.

Assume $m \geq 5$. In any case, $3 / m \geq 1805 / 1806 - 1 / 2$, so $m \leq 6$. Then $2 / l \geq 1805 / 1806 - 1 / 2 - 1 / 5$, so $l \leq 6$. Additionally, $l \geq 6$ (since $l > m$) and $1 / k \geq 1805 / 1806 - 1 / 2 - 1 / 5 - 1 / 6$, so $k \leq 7$, and $k l m n \leq 2 \cdot 6 \cdot 6 \cdot 7 = 504 < 1806$. From this, we similarly conclude that the quadruple $(k, l, m, n)$ is not optimal.

The only remaining case is $m=3$. Then $2 / l \geq 1805 / 1806 - 1 / 2 - 1 / 3$, so $l \leq 12$. If $l \geq 8$, then $1 / k \geq 1805 / 1806 - 1 / 2 - 1 / 3 - 1 / 8$, so $k \leq 24$. We get $k l m n \leq 2 \cdot 3 \cdot 12 \cdot 24 = 1728 < 1806$, which means the quadruple is not optimal. Therefore, $l=7$ and $k=43$, meaning the optimal quadruple was found at the beginning.

Answer: $(2,3,7,43)$.