Everyone knows that for an integer value $n$ there is a cyclic group, namely $Z(n)$. Hence for any integer $n$ there always is an abelian group. So why are we given exercises like:
Prove that the group of order $n$ (say $n=1225$) is an abelian group.
For every positive integer $n$ there is an abelian group of order $n$, namely the cyclic group of that order.
For some integers $n$, such as the prime numbers, this is the only group of that order (up to isomorphism). And all groups of that order are thus abelian.
For still some other integers, such as $4$, there are several groups of that order but all of them are abelian.
It is thus a meaningful question to ask if for some specific $n$ all groups of order $n$ are abelian.
For example one could ask:
Let $G$ be group of order $4$. Prove that $G$ is abelian.
This is not the same as.
Show that an abelain group of order $4$ exists.
This would be easy to answer for every $n$ as you remarked.
There may be abelian groups of all orders, but the question is asking you to show that any group of order $1225$ must be abelian.
Suppose $G$ is a group of order $1225$. $1225=35^2=5^27^2$. Sylow theory says that the number of subgroups $n_5$ of $G$ of order $5^2$ satisfies $n_5 \equiv 1$ mod $5$ and that $n_5 | 7^2$. The only solution to this is $n_5=1$, as $n_5 \in \{1,7,49\}$ and only $1 \equiv 1 \text{ mod } 5$ in this set. Similarly, $n_7=1$. So, by Sylow's theorems again, $G \cong A \times B$, where $A$ is a group of order $5^2$ and $B$ is a group of order $7^2$. As a group of order $p^2$ is always abelian for $p$ prime, and a direct product of abelian groups is abelian, $G$ must be abelian. $\square$