26. 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 $T$ with a side length of 1, so that no two squares have a common interior point
26. 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 $W$.

Lemma Proof 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: $\rho\left(M_{0}, l_{i}\right) \leqslant \frac{1}{2} (i=1,2,3,4)$, 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 the length of $l_{0} \leqslant MN$, and $MP = ME \triangleq \alpha, NP = NF \triangleq \beta$, 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 $\alpha + \beta = \frac{1}{2} - 2\alpha\beta \frac{1}{2}$. Take two squares $U, V$ with areas $\frac{S}{2}$ each, since $\frac{S}{2} > \frac{1}{4}$, according to the lemma, regardless of how squares $U, V$ are placed inside the unit square $T$, the center $O$ of $T$ must be a common interior point of $U, V$. When $S \leqslant \frac{1}{2}$, we can prove by analogy with the proof in Example 13 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. Thus, the maximum value of $S$ is $\frac{1}{2}$.