Two sums, each consisting of $n$ addends , are shown below:
$S = 1 + 2 + 3 + 4 + ...$
$T = 100 + 98 + 96 + 94 +...$ .
For what value of $n$ is it true that $S = T$ ?
1. **Identify the series and their properties:**
   - The first series \( S \) is an arithmetic series starting from 1 with a common difference of 1.
   - The second series \( T \) is an arithmetic series starting from 100 with a common difference of -2.

2. **Determine the sum of the first series \( S \):**
   - The sum of the first \( n \) natural numbers is given by:
     \[
     S = \sum_{k=1}^{n} k = \frac{n(n+1)}{2}
     \]

3. **Determine the sum of the second series \( T \):**
   - The \( n \)-th term of the series \( T \) can be expressed as:
     \[
     T_n = 100 + (n-1)(-2) = 100 - 2(n-1) = 102 - 2n
     \]
   - The sum of the first \( n \) terms of an arithmetic series is given by:
     \[
     T = \sum_{k=1}^{n} (102 - 2k) = n \left( \frac{100 + (102 - 2(n-1))}{2} \right) = n \left( \frac{100 + 102 - 2n + 2}{2} \right) = n \left( \frac{102 - 2n + 102}{2} \right) = n(101 - n)
     \]

4. **Set the sums equal to each other:**
   \[
   \frac{n(n+1)}{2} = n(101 - n)
   \]

5. **Simplify and solve for \( n \):**
   - First, cancel \( n \) from both sides (assuming \( n \neq 0 \)):
     \[
     \frac{n+1}{2} = 101 - n
     \]
   - Multiply both sides by 2 to clear the fraction:
     \[
     n + 1 = 202 - 2n
     \]
   - Combine like terms:
     \[
     n + 2n = 202 - 1
     \]
     \[
     3n = 201
     \]
   - Divide by 3:
     \[
     n = 67
     \]

The final answer is \( \boxed{67} \)