Minimum pumping length of (01)* Michael Sipser offers the definition:
The pumping lemma says that every regular language has a pumping length p, such that every string in the language can be pumped if it has length p or more. If p is a pumping length for language A, so is any length p′ ≥ p. The minimum pumping length for A is the smallest p that is a pumping length for A.
Now, (01)* in set notation is {€, 01,0101,010101....}
Taking minimum pumping length = 1, according to the definition, we have the statement if a string in the language has length 1 or more, it can be pumped.
This statement is true for all elements of the above mentioned set, so can the minimum pumping length be 1?
p.s. the minimum pumping length for (01)* has been asked here before but it doesn't answer my doubt that since the condition holds for minimum pumping length = 1, why is it not the answer?
The minimum pumping length $\ell$ of $(01)^*$ is $2$.
First of all notice that there are no words of length $1$ in the language.
Then, if you look at a formal definition of the pumping lemma for regular languages (what you wrote is still informal) you'll find that the length $|xy|$ of the prefix $xy$ of a word in the language that can be pumped must be $\le \ell$.
This allows you to prove that the minimum pumping length $\ell$ must be $2$ by showing:

That there is a DFA for $(01)^*$ that uses at most $2$ states, implying $\ell \le 2$. This DFA is trivial to obtain.
That $\ell=1$ is not a valid pumping length. This is true since $|y|$ must be positive and, together with $|xy| \le \ell$, this implies $|y|=1$. In other words $x = \epsilon$ and $y=0$. It is easy to obtain a word not in the language by repeating $y$ either $0$ or at least $2$ times.