Let $S$ and $T$ be non-empty, finite sets of positive integers. We say that $a\in\mathbb{N}$ is \emph{good} for $b\in\mathbb{N}$ if $a\geq\frac{b}{2}+7$. We say that an ordered pair $\left(a,b\right)\in S\times T$ is \emph{satisfiable} if $a$ and $b$ are good for each other.

A subset $R$ of $S$ is said to be \emph{unacceptable} if there are less than $\left|R\right|$ elements $b$ of $T$ with the property that there exists $a \in R$ such that $\left(a,b\right)$ is satisfiable. If there are no unacceptable subsets of $S$, and $S$ contains the elements $14$, $20$, $16$, $32$, $23$, and $31$, compute the smallest possible sum of elements of $T$ given that $\left|T\right|\geq20$. 

[i]Proposed by Tristan Shin[/i]
To solve this problem, we need to ensure that for every subset \( R \) of \( S \), there are at least \( |R| \) elements in \( T \) such that each element \( b \in T \) is good for some \( a \in R \). We start by analyzing the given elements of \( S \) and determining the intervals of \( b \) that are good for each \( a \).

1. **Determine the intervals for each \( a \in S \):**
   - For \( a = 14 \):
     \[
     a \geq \frac{b}{2} + 7 \implies 14 \geq \frac{b}{2} + 7 \implies 14 - 7 \geq \frac{b}{2} \implies 7 \geq \frac{b}{2} \implies b \leq 14
     \]
     So, \( b \in [1, 14] \).

   - For \( a = 16 \):
     \[
     a \geq \frac{b}{2} + 7 \implies 16 \geq \frac{b}{2} + 7 \implies 16 - 7 \geq \frac{b}{2} \implies 9 \geq \frac{b}{2} \implies b \leq 18
     \]
     So, \( b \in [1, 18] \).

   - For \( a = 20 \):
     \[
     a \geq \frac{b}{2} + 7 \implies 20 \geq \frac{b}{2} + 7 \implies 20 - 7 \geq \frac{b}{2} \implies 13 \geq \frac{b}{2} \implies b \leq 26
     \]
     So, \( b \in [1, 26] \).

   - For \( a = 23 \):
     \[
     a \geq \frac{b}{2} + 7 \implies 23 \geq \frac{b}{2} + 7 \implies 23 - 7 \geq \frac{b}{2} \implies 16 \geq \frac{b}{2} \implies b \leq 32
     \]
     So, \( b \in [1, 32] \).

   - For \( a = 31 \):
     \[
     a \geq \frac{b}{2} + 7 \implies 31 \geq \frac{b}{2} + 7 \implies 31 - 7 \geq \frac{b}{2} \implies 24 \geq \frac{b}{2} \implies b \leq 48
     \]
     So, \( b \in [1, 48] \).

   - For \( a = 32 \):
     \[
     a \geq \frac{b}{2} + 7 \implies 32 \geq \frac{b}{2} + 7 \implies 32 - 7 \geq \frac{b}{2} \implies 25 \geq \frac{b}{2} \implies b \leq 50
     \]
     So, \( b \in [1, 50] \).

2. **Select the smallest elements of \( T \) that cover all intervals:**
   - We need to ensure that \( T \) has at least 20 elements and that for any subset \( R \) of \( S \), there are enough \( b \) values in \( T \) to satisfy the conditions.
   - We start by selecting the smallest elements that cover the intervals for each \( a \):
     \[
     T = \{1, 2, 3, \ldots, 14, 15, 16, 17, 18, 19, 20\}
     \]

3. **Verify the sum of the selected elements:**
   - The sum of the first 20 positive integers is:
     \[
     \sum_{i=1}^{20} i = \frac{20 \cdot (20 + 1)}{2} = 210
     \]

The final answer is \( \boxed{210} \).