Let $m$ be the product of the first 100 primes, and let $S$ denote the set of divisors of $m$ greater than 1 (hence $S$ has exactly $2^{100} - 1$ elements).  We wish to color each element of $S$ with one of $k$ colors such that
$\ \bullet \ $ every color is used at least once; and
$\ \bullet \ $ any three elements of $S$ whose product is a perfect square have exactly two different colors used among them.
Find, with proof, all values of $k$ for which this coloring is possible.
1. **Restate the problem in terms of vector space:**
   Let \( V \) denote the set of vectors of length \( n \) whose entries are either \( 0 \) or \( 1 \), with addition taken modulo \( 2 \). Each number \( s \in S \) can be identified with a vector in \( V \) by taking the \( i \)-th coordinate to be nonzero if \( s \) is divisible by the \( i \)-th prime. The condition that three elements have a square product is equivalent to the corresponding three vectors having sum zero.

2. **Rephrase the problem:**
   The \( 2^n - 1 \) nonzero elements of \( V \) need to be colored with \( k \) colors (each color used at least once) such that if \( a, b, c \in V \) are nonzero and satisfy \( a + b + c = \mathbf{0} \), then exactly two different colors are used among them.

3. **Example of valid coloring with exactly \( n \) colors:**
   Given each \( a \in V \), if the first \( 1 \) is in the \( i \)-th coordinate, color \( a \) by the \( i \)-th color.

4. **Proof that any coloring uses exactly \( n \) colors:**
   We prove this by induction on \( n \ge 1 \). The base case \( n = 1 \) is obvious. For the inductive step, assume the statement is true for \( n-1 \).

5. **Key claim:**
   If \( a, b, c \) are the same color and \( a + b + c \neq \mathbf{0} \), then \( a + b + c \) is also this color.
   - **Proof:** Suppose not. Let \( \color{red} a \), \( \color{red} b \), \( \color{red} c \) be "red" and \( \color{blue} a + b + c \) be "blue". Then,
     \[
     \underbrace{\color{red} b}_{\text{red}} + \underbrace{\color{red} c}_{\text{red}} = \underbrace{\color{red} a}_{\text{red}} + (\underbrace{\color{blue} a+b+c}_{\text{blue}})
     \]
     implies \( \color{blue} b+c \) must be blue. Similarly, \( \color{blue} c+a \) and \( \color{blue} a+b \) are blue. This is a contradiction because these three elements sum to zero. \( \blacksquare \)

6. **Implication of the key claim:**
   Pick any color. The number of times it is used is either \( 1 \) or an even number.
   - **Proof:** Call the color "red". Suppose there are at least two red elements, say \( \color{red} a \) and \( \color{red} b \). Note that
     \[
     c \coloneqq {\color{red}a} + {\color{red}b} \neq \mathbf{0}
     \]
     is not red. By the key claim, if \( x \) is red, then so is \( x + c \), and conversely, if \( x + c \) is red, then so is \( x \). It follows that the elements of red split into pairs which sum to \( c \), so there must be an even number of them. \( \blacksquare \)

7. **Conclusion from the above claim:**
   Since \( 2^n - 1 \) is odd, some color must appear exactly once. Suppose that element \( v \) has a unique color, and WLOG \( v \) has a \( 1 \) in the \( n \)-th component. Then, \( x \) and \( x + v \) have the same color for all \( x \neq \mathbf{0} \); moreover, exactly one of these two has a \( 0 \) in the \( n \)-th component.

8. **Reduction to \( n-1 \):**
   This means we can quotient out by \( v \). Take the \( 2^{n-1} - 1 \) pairs \( \{x, x+v\} \) described above. For each pair, take the vector with a \( 0 \) in the \( n \)-th component, and delete this component to get a vector of length \( n-1 \). If we do this process, we obtain one (nonzero) vector of length \( n-1 \) from each pair, together with a valid coloring with exactly one fewer color (because the unique color for \( v \) is the only one that disappeared). By the induction hypothesis, exactly \( n-1 \) colors were left. So there were exactly \( n \) colors to begin with.

9. **Conclusion:**
   The only possible value of \( k \) is \( k = n \).

The final answer is \( \boxed{ k = n } \).