The points $A,B,C,D,T_1$ determine $4$ triangles: $t_1=\triangle{ABT_1};t_2=\triangle{BCT_1};t_3=\triangle{CDT_1};t_4=\triangle{DAT_1}$ with the properties: $\bullet I(t_i)\cap I(t_j)=\phi;\;\forall i,j\in\{1,2,3,4\},i\ne j$, where $I(t_i)$ denotes the interior of the triangle $t_i$; $\bullet \sum_{i=1}^4 \mathcal{A}(t_i)=\mathcal{A}(ABCD)=400$, where $\mathcal{A}(t_i)$ denotes the area of the triangle $t_i$ and $\mathcal{A}(ABCD)$ the area of the square $ABCD$. WLOG, we can consider $T_2\in I(t_1)$. The point $T_2$ splits the triangle $\triangle{ABT_1}$ into the triangles $t_5=\triangle{ABT_2};t_6=\triangle{AT_1T_2};t_7=\triangle{BT_1T_2}$ with the properties: $\bullet I(t_5)\cap I(t_6)=I(t_5)\cap I(t_7)=I(t_5)\cap I(t_7)=\phi$; $\bullet \mathcal{A}(t_5)+\mathcal{A}(t_6)+\mathcal{A}(t_7)=\mathcal{A}(t_1)$. Results: Instead the triangle $t_1$ appear $3$ new triangles $t_5;t_6;t_7$ and $\sum_{i=2}^7 \mathcal{A}(t_i)=\mathcal{A}(ABCD)=400$, hence the square $ABCD$ is a partition of $6$ triangles. With this construction, the total number of triangles which form a partition of $ABCD$ increases with $2$ comparative with the previous construction. Similarly, for each $k\in\{2,3,\dots,1999\}$: Assume the square $ABCD$ is a partition of $n_{k}$ triangles with the vertices in the set $\{A,B,C,D,T_1,T_2,\dots,T_{k-1},T_{k}\}$. Then exist the distinct points $M,N,P\in\{A,B,C,D,T_1,T_2,\dots,T_{k-1},T_{k}\}$ such that $T_{k+1}\in I(\triangle{MNP})$ and $\triangle{MNP}$ is the smallest triangle with this property; Instead the triangle $\triangle{MNP}$ appear $3$ new triangles $t'=\triangle{MNT_{k+1}};t''=\triangle{MPT_{k+1}};t'''=\triangle{NPT_{k+1}}$ with $\mathcal{A}(MNP)=\mathcal{A}(MNT_{k+1})+\mathcal{A}(MPT_{k+1})+\mathcal{A}(NPT_{k+1})$ and the square $ABCD$ is a partition of $n_{k+1}=n_k+2$ triangles with the vertices in the set $\{A,B,C,D,T_1,T_2,\dots,T_{k},T_{k+1}\}$. By induction, $n_k=2k+2,\;\forall k\in\{1,2,\dots,2000\}$. Hence, for $k=2000$: the square $ABCD$ is a partition of $4002$ triangles with the vertices in $S$ and the minimum of the areas of these triangles is $\mathcal{A}(t_{\text{min}})\le \dfrac{\mathcal{A}}{4002}<\dfrac{400}{4000}=\dfrac{1}{10}$.

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