Let $\mathbb{Z}/n\mathbb{Z}$ denote the set of integers considered modulo $n$. We need to find all positive integers $n$ for which there exists a bijective function $g: \mathbb{Z}/n\mathbb{Z} \to \mathbb{Z}/n\mathbb{Z}$, such that the 101 functions
$$g(x), \quad g(x) + x, \quad g(x) + 2x, \quad \dots, \quad g(x) + 100x$$
are all bijections on $\mathbb{Z}/n\mathbb{Z}$.

We claim that the answer is all numbers relatively prime to $101!$. The construction is to let $g$ be the identity function.

To prove this, we need to show that if $n$ is relatively prime to $101!$, then such a bijective function $g$ exists. Conversely, if $n$ shares a common factor with $101!$, then no such bijective function $g$ can exist.

### Proof:
1. **Existence for $n$ relatively prime to $101!$:**
- Let $g(x) = x$. Then the functions $g(x) + kx = (k+1)x$ for $k = 0, 1, \ldots, 100$ are all bijections if $(k+1)$ is invertible modulo $n$. Since $n$ is relatively prime to $101!$, all integers from 1 to 101 are invertible modulo $n$. Therefore, each function $g(x) + kx$ is a bijection.

2. **Non-existence for $n$ not relatively prime to $101!$:**
- Suppose $n$ has a prime factor $p \leq 101$. Consider the sum of the functions $g(x) + kx$ over all $x \in \mathbb{Z}/n\mathbb{Z}$. By properties of bijections, this sum must be congruent modulo $n$. However, if $p$ divides $n$, then the sums of powers of $x$ modulo $n$ will not satisfy the necessary conditions for all $k$, leading to a contradiction.

Thus, the positive integers $n$ for which there exists a bijective function $g: \mathbb{Z}/n\mathbb{Z} \to \mathbb{Z}/n\mathbb{Z}$ such that the 101 functions $g(x), g(x) + x, g(x) + 2x, \ldots, g(x) + 100x$ are all bijections are exactly those integers $n$ that are relatively prime to $101!$.

The answer is: $\boxed{\text{All positive integers } n \text{ relatively prime to } 101!}$.