Let $\mathcal{S}$ be the set of all positive integers which are both a multiple of $3$ and have at least one digit that is a $1$. For example, $123$ is in $\mathcal{S}$ and $450$ is not. The probability that a randomly chosen $3$-digit positive integer is in $\mathcal{S}$ can be written as $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

[i]Proposed by GammaZero[/i]
To solve this problem, we need to determine the probability that a randomly chosen 3-digit positive integer is in the set $\mathcal{S}$, which consists of numbers that are multiples of 3 and have at least one digit that is 1. We will use the Principle of Inclusion-Exclusion (PIE) to count the numbers in $\mathcal{S}$.

1. **Total number of 3-digit integers:**
   The range of 3-digit integers is from 100 to 999. Therefore, there are:
   \[
   999 - 100 + 1 = 900 \text{ three-digit integers.}
   \]

2. **Counting multiples of 3:**
   A number is a multiple of 3 if the sum of its digits is divisible by 3. We need to count the 3-digit numbers that are multiples of 3.
   - The smallest 3-digit multiple of 3 is 102.
   - The largest 3-digit multiple of 3 is 999.
   - These numbers form an arithmetic sequence with the first term \(a = 102\) and common difference \(d = 3\).

   The number of terms in this sequence is given by:
   \[
   n = \frac{999 - 102}{3} + 1 = 300.
   \]
   So, there are 300 multiples of 3 among the 3-digit numbers.

3. **Counting multiples of 3 with at least one digit being 1:**
   We will use PIE to count the numbers that are multiples of 3 and have at least one digit equal to 1.

   - **Case 1: 1 is the first digit (hundreds place):**
     If the first digit is 1, the number is of the form \(1xy\), where \(x\) and \(y\) are digits. The sum \(1 + x + y\) must be divisible by 3. There are 10 choices for \(x\) and 10 choices for \(y\), but we need to count only those combinations where \(1 + x + y\) is divisible by 3.
     - For each \(x\), there are 3 values of \(y\) that make \(1 + x + y\) divisible by 3.
     - Therefore, there are \(10 \times 3 = 30\) such numbers.

   - **Case 2: 1 is the second digit (tens place):**
     If the second digit is 1, the number is of the form \(x1y\). Similar to the previous case, we need \(x + 1 + y\) to be divisible by 3.
     - For each \(x\), there are 3 values of \(y\) that make \(x + 1 + y\) divisible by 3.
     - Therefore, there are \(10 \times 3 = 30\) such numbers.

   - **Case 3: 1 is the third digit (units place):**
     If the third digit is 1, the number is of the form \(xy1\). We need \(x + y + 1\) to be divisible by 3.
     - For each \(x\), there are 3 values of \(y\) that make \(x + y + 1\) divisible by 3.
     - Therefore, there are \(10 \times 3 = 30\) such numbers.

   - **Subtracting overcounted cases:**
     We have overcounted numbers where two digits are 1. We need to subtract these cases.
     - If two digits are 1, the number is of the form \(11x\), \(1x1\), or \(x11\). In each case, the remaining digit \(x\) must be such that the sum of the digits is divisible by 3.
     - For each of these forms, there are 3 values of \(x\) that make the sum divisible by 3.
     - Therefore, there are \(3 + 3 + 3 = 9\) such numbers.

   - **Adding back the case where all three digits are 1:**
     The number 111 is a multiple of 3 and has all digits as 1. We have subtracted this case three times, so we need to add it back once.

   Combining all these, we get:
   \[
   30 + 30 + 30 - 9 + 1 = 82.
   \]

4. **Calculating the probability:**
   The probability that a randomly chosen 3-digit number is in $\mathcal{S}$ is:
   \[
   \frac{82}{900}.
   \]
   Simplifying this fraction, we get:
   \[
   \frac{82}{900} = \frac{41}{450}.
   \]

5. **Finding \(m + n\):**
   The fraction \(\frac{41}{450}\) is in simplest form, where \(m = 41\) and \(n = 450\). Therefore:
   \[
   m + n = 41 + 450 = 491.
   \]

The final answer is \(\boxed{491}\).