One, (20 points) Given prime numbers $p$ and $q$ such that the algebraic expressions $\frac{2 p+1}{q}$ and $\frac{2 q-3}{p}$ are both natural numbers. Try to find the value of $p^{2} q$.

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Let $p \geqslant q$, then we have $1 \leqslant \frac{2 q-3}{p}<2$.
Thus, it can only be $\frac{2 q-3}{p}=1$, which means $p=2 q-3$.
At this point, $\frac{2 p+1}{q}=\frac{4 q-5}{q}=4-\frac{5}{q}$.
To make $\frac{2 p+1}{q}$ a natural number, it can only be $q=5$, hence $p=7$.

Now, let $p<q$, in this case, $1 \leqslant \frac{2 p+1}{q}<3$, which can be divided into the following two scenarios:
(1) $\frac{2 p+1}{q}=1, q=2 p+1$.
At this point, $\frac{2 q-3}{p}=\frac{4 p-1}{p}$, which gives $p=1$, which does not meet the requirements.
(2) $\frac{2 p+1}{q}=2$, i.e., $2 p+1=2 q$, the left side is odd, the right side is even, a contradiction.
Therefore, the only $p$ and $q$ that satisfy the conditions are $p=7$ and $q=5$.
Thus, $p^{2} q=49 \times 5=245$.