Each of the first $150$ positive integers is painted on a different marble, and the $150$ marbles are placed in a bag. If $n$ marbles are chosen (without replacement) from the bag, what is the smallest value of $n$ such that we are guaranteed to choose three marbles with consecutive numbers?
To determine the smallest value of \( n \) such that we are guaranteed to choose three marbles with consecutive numbers from the first 150 positive integers, we need to consider the worst-case scenario where we avoid picking three consecutive numbers as long as possible.

1. **Identify the pattern to avoid three consecutive numbers:**
   - We can choose numbers in such a way that we avoid picking three consecutive numbers. For example, we can pick numbers in the following pattern:
     \[
     1, 2, 4, 5, 7, 8, 10, 11, \ldots, 148, 149
     \]
     or:
     \[
     2, 3, 5, 6, 8, 9, \ldots, 149, 150
     \]

2. **Count the numbers in the pattern:**
   - In each pattern, we are effectively skipping every third number. To see how many numbers we can pick without having three consecutive numbers, we can count the numbers in one of these patterns.
   - Consider the first pattern: \( 1, 2, 4, 5, 7, 8, \ldots, 148, 149 \).
   - We can see that every block of three numbers (e.g., \( 1, 2, 3 \)) allows us to pick two numbers. Therefore, for every three numbers, we can pick two.

3. **Calculate the total number of numbers we can pick:**
   - There are 150 numbers in total. We divide these into blocks of three:
     \[
     \left\lfloor \frac{150}{3} \right\rfloor = 50 \text{ blocks}
     \]
   - In each block, we can pick 2 numbers, so:
     \[
     50 \times 2 = 100 \text{ numbers}
     \]

4. **Determine the smallest value of \( n \):**
   - If we pick 100 numbers, we can avoid having three consecutive numbers.
   - However, if we pick one more number (i.e., 101 numbers), we are forced to have at least one set of three consecutive numbers because we have exhausted all possible ways to avoid three consecutive numbers.

Therefore, the smallest value of \( n \) such that we are guaranteed to choose three marbles with consecutive numbers is \( 101 \).

The final answer is \(\boxed{101}\).