Let $f(n) = \sum^n_{d=1} \left\lfloor \frac{n}{d} \right\rfloor$ and $g(n) = f(n) -f(n - 1)$. For how many $n$ from $1$ to $100$ inclusive is $g(n)$ even?
1. **Define the function \( f(n) \):**
   \[
   f(n) = \sum_{d=1}^n \left\lfloor \frac{n}{d} \right\rfloor
   \]
   This function sums the floor of the division of \( n \) by each integer from 1 to \( n \).

2. **Express \( f(n) \) in terms of the number of divisors:**
   We claim that:
   \[
   f(n) = \sum_{k=1}^n \tau(k)
   \]
   where \( \tau(k) \) is the number of divisors of \( k \). To understand why this is true, consider the following steps:
   \[
   \sum_{k=1}^n \tau(k) = \sum_{k=1}^n \sum_{j \mid k} 1
   \]
   Here, \( j \mid k \) means \( j \) is a divisor of \( k \). We can rewrite this as:
   \[
   \sum_{k=1}^n \sum_{j \mid k} 1 = \sum_{j=1}^n \sum_{k=1}^n [j \mid k]
   \]
   where \( [j \mid k] \) is 1 if \( j \) divides \( k \) and 0 otherwise. This can be further simplified to:
   \[
   \sum_{j=1}^n \sum_{k=1}^n [j \mid k] = \sum_{j=1}^n \left\lfloor \frac{n}{j} \right\rfloor
   \]
   Therefore, we have:
   \[
   f(n) = \sum_{j=1}^n \left\lfloor \frac{n}{j} \right\rfloor
   \]

3. **Define the function \( g(n) \):**
   \[
   g(n) = f(n) - f(n-1)
   \]
   Using the expression for \( f(n) \), we get:
   \[
   g(n) = \sum_{k=1}^n \tau(k) - \sum_{k=1}^{n-1} \tau(k) = \tau(n)
   \]
   Thus, \( g(n) = \tau(n) \), which is the number of divisors of \( n \).

4. **Determine when \( g(n) \) is even:**
   \( g(n) \) is even if \( \tau(n) \) is even. The number of divisors \( \tau(n) \) is even if and only if \( n \) is not a perfect square. This is because a perfect square has an odd number of divisors (since one of the divisors is repeated).

5. **Count the number of non-perfect squares from 1 to 100:**
   The perfect squares between 1 and 100 are \( 1, 4, 9, 16, 25, 36, 49, 64, 81, 100 \). There are 10 perfect squares in this range.

   Therefore, the number of non-perfect squares from 1 to 100 is:
   \[
   100 - 10 = 90
   \]

Conclusion:
\[
\boxed{90}
\]