Given Conditions:
1. When $n$ is divided by 3 , the remainder is 1 .

This can be written as: $\(n \equiv 1(\bmod 3)\)$
So, possible values for $n$ are $\(1,4,7,10,13,16,19,22,25, \ldots\)$
2. When $n$ is divided by 4 , the remainder is 2 .

This can be written as: $\(n \equiv 2(\bmod 4)\)$
So, possible values for $n$ are $\(2,6,10,14,18,22,26, \ldots\)$
Finding the Smallest Common Value:
We are looking for a number that appears in both lists.
By comparing the two lists, we can see that 10 is the first common number.
Let's check if 10 satisfies both conditions:
- $\(10 \div 3=3\)$ remainder 1 (Satisfied)
- $\(10 \div 4=2\)$ remainder 2 (Satisfied)

Since 10 is the smallest such positive integer, we can find other values of $n$ by adding multiples of the least common multiple (LCM) of 3 and 4.
$\(\operatorname{LCM}(3,4)=12\)$.
So, any integer $n$ that satisfies these conditions will be of the form $10+12 k$, where $k$ is an integer.
For example, if $k=1, n=10+12=22$.
- $\(22 \div 3=7\)$ remainder 1
- $\(22 \div 4=5\)$ remainder 2

This also works.
Finding the Remainder when $n$ is Divided by 12 :
Since $n$ can be written in the form $10+12 k$, when $n$ is divided by 12 , the $12 k$ part will be perfectly divisible by 12 , leaving no remainder from that part. The remainder will come solely from the 10.

Therefore, the remainder when $n$ is divided by 12 is 10 .
The final answer is 10 .