20 Find the largest $S$, such that any finite number of small squares with a total area of $S$ can always be placed inside a square $T$ of side length 1, so that no two of the small squares have a common interior point.
Lemma: If a square $W$ is covered by a unit square $T$, and if the area of $W$ is greater than $\frac{1}{4}$ (side length greater than $\frac{1}{2}$), then the center $O$ of $T$ must be inside $W$.

Proof of the Lemma: Let $\rho(M, l)$ denote the distance from point $M$ to line $l$. Since the distance between any pair of opposite sides of square $W$ is greater than $\frac{1}{2}$, if the four sides of $W$ are lines $l_{1}, l_{2}, l_{3}, l_{4}$, and point $M_{0}$ satisfies:
$$
\rho\left(M_{0}, l_{i}\right) \leqslant \frac{1}{2} \quad (i=1,2,3,4)
$$
then point $M_{0}$ must be inside $W$.
(1) When the sides of square $W$ are parallel to the sides of the unit square $T$, the center $O$ of $T$ clearly satisfies $\rho\left(O, l_{i}\right) \leqslant \frac{1}{2}, i=1,2,3,4$, so $O$ must be inside $W$.
(2) If the sides of square $W$ are not parallel to the sides of the unit square $T$, let the incircle $\odot O$ of square $T$ touch the sides of $T$ at four points $E, F, G, H$, with radius $r=\frac{1}{2}$. For any line $l$ that intersects or is tangent to $\odot O$, we have $\rho(O, l) \leqslant \frac{1}{2}$. Suppose one side $l_{0}$ of square $W$ does not intersect $\odot O$, and $l_{0}$ is contained within the curvilinear triangle $AEF$. Translate $l_{0}$ towards $\odot O$ to $l'$, such that $l'$ is tangent to $\odot O$ at point $P$. Let $l'$ intersect $AB$ and $AD$ at points $M$ and $N$, respectively, then $l_{0} \leqslant MN$, and $MP = ME \cong \alpha, NP = NF \cong \beta$, so
$$
AM = \frac{1}{2} - \alpha, AN = \frac{1}{2} - \beta, MN = \alpha + \beta, \alpha \beta \neq 0,
$$
In the right triangle $\triangle AMN$, $MN^2 = AM^2 + AN^2$, i.e.,
$$
(\alpha + \beta)^2 = \left(\frac{1}{2} - \alpha\right)^2 + \left(\frac{1}{2} - \beta\right)^2,
$$
so
$$
\alpha + \beta = \frac{1}{2} - 2\alpha\beta \leq \frac{1}{2}.
$$
Consider two squares $U$ and $V$ with area $\frac{S}{2}$ each, since $\frac{S}{2} > \frac{1}{4}$, according to the lemma, regardless of how $U$ and $V$ are placed inside the unit square $T$, the center $O$ of $T$ must be a common interior point of $U$ and $V$.

When $S \leq \frac{1}{2}$, we will prove that any finite number of squares with a total area of $\frac{1}{2}$ can be placed inside a unit square such that no two small squares have a common interior point.

To this end, arrange all the small squares in order of decreasing side length, denoted as $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{n}$. Let $\sigma_{k} = \sum_{i=1}^{k} a_{i}, k=1,2, \cdots, n$. If $\sigma_{k_{1}} \leqslant 1$ and $\sigma_{k_{1}+1} > 1$, then redefine $\sigma_{k_{1}} = r_{1}$; if $\sigma_{k_{2}} - r_{1} \leqslant 1$ and $\sigma_{k_{2}+1} - r_{1} > 1$, then redefine $\sigma_{k_{2}} - r_{1} = r_{2}, \cdots$. This divides the sequence of squares into segments: $r_{1}, r_{2}, r_{3}, \cdots$; let the side length of the first square in the $i$-th segment be $h_{i}, i=1,2, \cdots$, then $h_{1} \geqslant h_{2} \geqslant \cdots$, and
$$
r_{i} \leqslant 1, r_{i} + h_{i+1} > 1, i=1,2, \cdots,
$$
which implies
$$
r_{i} > 1 - h_{i+1} \geqslant 1 - h_{1}, i=1,2, \cdots.
$$
Since $a_{1}^2 + a_{2}^2 + \cdots + a_{n}^2 = \frac{1}{2}$ and $a_{1} = h_{1}$, we have
$$
\begin{aligned}
\frac{1}{2} - h_{1}^2 = & a_{2}^2 + a_{3}^2 + \cdots + a_{n}^2 \geqslant h_{2}(r_{1} + h_{2} - h_{1}) \\
& + h_{3}(r_{2} + h_{3} - h_{2}) + h_{4}(r_{3} + h_{4} - h_{3}) + \cdots \\
\geqslant & h_{2}(1 - h_{1}) + h_{3}(1 - h_{2}) + h_{4}(1 - h_{3}) + \cdots \\
\geqslant & h_{2}(1 - h_{1}) + h_{3}(1 - h_{1}) + h_{4}(1 - h_{1}) + \cdots \\
= & (1 - h_{1})(h_{2} + h_{3} + h_{4} + \cdots).
\end{aligned}
$$
(Here, $r_{i} + h_{i+1} - h_{i}$ represents the sum of the side lengths of a group of squares from the $i$-th segment, excluding the first square and including the first square of the next segment, with the height calculated as the smallest side length $h_{i+1}$, so the area of this group of squares is at least $h_{i+1}(r_{i} + h_{i+1} - h_{i}), i=1, 2, \cdots$). Adding $h_{1}(1 - h_{1})$ to both sides, we get
$$
\frac{1}{2} + h_{1} - 2h_{1}^2 \geqslant (1 - h_{1})(h_{1} + h_{2} + h_{3} + \cdots).
$$
Noting that $1 - h_{1} \geqslant \frac{1}{2} + h_{1} - 2h_{1}^2$, we have
$$
1 - h_{1} \geqslant (1 - h_{1})(h_{1} + h_{2} + \cdots + h_{n}),
$$
and dividing both sides by the positive number $1 - h_{1}$, we get
$$
h_{1} + h_{2} + h_{3} + \cdots + h_{n} \leqslant 1.
$$
Now, take the squares in segments of lengths $r_{1}, r_{2}, r_{3}, \cdots$, and pack all the squares in the $i$-th segment into a rectangle $E_{i}$ with length $r_{i}$ and height $h_{i}$, $i=1, 2, \cdots$. Then place the rectangles $E_{1}, E_{2}, E_{3}, \cdots$ one on top of the other. Since each rectangle has a length $r_{i} \leqslant 1$ and the total height of all rectangles $E_{1}, E_{2}, E_{3}, \cdots$ is $h_{1} + h_{2} + h_{3} + \cdots \leqslant 1$, all the small squares can be placed inside the unit square. Thus, the maximum value of $S$ is $\frac{1}{2}$.