Throughout this solution, we use the fact that if $N$ is a positive integer with $N>1$ and $N$ has prime factorization $p_{1}^{a_{1}} p_{2}^{a_{2}} \cdots p_{m}^{a_{m}}$ for some distinct prime numbers $p_{1}, p_{2}, \ldots, p_{m}$ and positive integers $a_{1}, a_{2}, \ldots, a_{m}$, then the number of positive divisors of $N$ including 1 and $N$ is equal to $(1+a_{1})(1+a_{2}) \cdots(1+a_{m})$. We are told that $N$ is a positive multiple of 2024. Now, $2024=8 \cdot 253=2^{3} \cdot 11 \cdot 23$. This means that $N$ has at least 3 prime factors (namely 2,11 and 23) and that at least one of these prime factors has an exponent of at least 3. Let $D$ be the number of positive divisors that $N$ has. We are told that $100<D<110$. Since $D=(1+a_{1})(1+a_{2}) \cdots(1+a_{m})$ and $N$ has at least 3 prime factors, then $D$ is a positive integer that can be written as the product of at least 3 positive integers each greater than 2. $D$ cannot equal $101,103,107$, or 109, since each of these is prime (and so cannot be written as the product of 3 integers each at least 2). $D$ also cannot equal 106 because $106=2 \cdot 53$ (both 2 and 53 are prime), which means that 106 cannot be written as the product of three integers each greater than 1. The possible values of $D$ that remain are $102,104,105,108$. We note that $102=2 \cdot 3 \cdot 17$ and $104=2^{3} \cdot 13$ and $105=3 \cdot 5 \cdot 7$ and $108=2^{2} \cdot 3^{3}$. Case 1: $D=102$ Since the prime factors of $N$ include at least 2,11 and 23, then the prime factorization of $N$ includes factors of $2^{a}, 11^{b}$ and $23^{c}$ for some positive integers $a, b$ and $c$ with $a \geq 3$. If a fourth prime power $p^{e}$ was also a factor of $N$, then $D$ would be divisible by $(1+a)(1+b)(1+c)(1+e) \cdot(D$ could have more factors if $N$ had more prime factors.) Since $D=102=2 \cdot 3 \cdot 17$ has only 3 prime factors, it cannot be written as the product of 4 integers each greater than 1. Thus, $N$ cannot have a fourth prime factor. This means that $N=2^{a} 11^{b} 23^{c}$, which gives $D=(1+a)(1+b)(1+c)=2 \cdot 3 \cdot 17$. This means that $1+a, 1+b$ and $1+c$ are equal to 2,3 and 17, in some order, and so $a, b$ and $c$ are equal to 1,2 and 16, in some order. For $D$ to be as small as possible, the largest exponent goes with the smallest prime, the next largest exponent with the next smallest prime, and so on. (Can you see why this makes $N$ as small as possible?) Therefore, the smallest possible value of $N$ in this case is $N=2^{16} 11^{2} 23^{1}=182386688$. Case 2: $D=105$ Using a similar argument, we can determine that $N=2^{a} 11^{b} 23^{c}$ with $1+a, 1+b$ and $1+c$ equal to 3,5 and 7 in some order, meaning that $a, b$ and $c$ equal $2,4,6$ in some order. Therefore, the minimum value of $N$ is this case is $N=2^{6} 11^{4} 23^{2}=495685696$. Case 3: $D=104$ Since $D=2^{3} \cdot 13$ has 4 prime factors, then $N$ cannot have more than 4 prime factors. (If $N$ had 5 or more prime factors, then the product equal to $D$ would include at least 5 integers, each at least 2.) Therefore, $N=2^{a} 11^{b} 23^{c}$ and $D=(1+a)(1+b)(1+c)$, or $N=2^{a} 11^{b} 23^{c} p^{e}$ for some prime $p \neq 2,11,23$ and $D=(1+a)(1+b)(1+c)(1+e)$. This means that $(1+a)(1+b)(1+c)=2^{3} \cdot 13$ or $(1+a)(1+b)(1+c)(1+e)=2^{3} \cdot 13$. In the case that $N$ has three prime factors, we note that $104=26 \cdot 2 \cdot 2=13 \cdot 4 \cdot 2$ are the only two ways of writing 104 as the product of 3 integers each of which is at least 2. These give corresponding minimum values of $N=2^{25} \cdot 11 \cdot 23=8489271296$ and $N=2^{12} \cdot 11^{3} \cdot 23=125390848$. In the case that $N$ has four prime factors, then $(1+a)(1+b)(1+c)(1+e)=2 \cdot 2 \cdot 2 \cdot 13$ means that $a, b, c$ and $e$ are $1,1,1,12$ in some order. This in turn means that the corresponding smallest possible value of $N$ is $N=2^{12} \cdot 3 \cdot 11 \cdot 23=3108864$. We note here that the prime power $p^{e}$ has become $3^{1}$ in order to minimize both $p$ (since $p>2$ ) and its exponent. Case 4: $D=108$ Since $D=2^{2} \cdot 3^{3}$ has 5 prime factors, then $N$ cannot have more than 5 prime factors. If $N$ has 5 prime factors, then we need to use the factorization $D=2 \cdot 2 \cdot 3 \cdot 3 \cdot 3$. This is not possible, however, because the power $2^{a}$ must have $a \geq 3$ which would mean that one of the five factors of $D$ would have to be at least 4. If $N$ has 4 prime factors, then $D$ must be partitioned as $9 \cdot 3 \cdot 2 \cdot 2$ or $6 \cdot 3 \cdot 3 \cdot 2$ or $4 \cdot 3 \cdot 3 \cdot 3$. (Since two of the prime factors have to be combined, either two 2 s , two 3 s , or a 2 and a 3 are combined.) These give minimum values of $N=2^{8} \cdot 3^{2} \cdot 11 \cdot 23=582912$ and $N=2^{5} \cdot 3^{2} \cdot 11^{2} \cdot 23=801504$ and $N=2^{3} \cdot 3^{2} \cdot 11^{2} \cdot 23^{2}=4408648$. If $N$ has 3 prime factors, then we must use one of the factorizations $D=27 \cdot 2 \cdot 2$ or $D=18 \cdot 3 \cdot 2$ or $D=12 \cdot 3 \cdot 3$ or $D=9 \cdot 4 \cdot 3$ or $D=6 \cdot 6 \cdot 3$. These gives corresponding minimum values $N=2^{26} \cdot 11 \cdot 23=16978542592$, $N=2^{17} \cdot 11^{2} \cdot 23=364773376$, $N=2^{11} \cdot 11^{2} \cdot 23^{2}=131090432$, $N=2^{8} \cdot 11^{3} \cdot 23^{2}=180249344$, $N=2^{5} \cdot 11^{5} \cdot 23^{2}=2726271328$. Combining Cases 1 through 4, the minimum possible value of $N$ is 582912. The sum of the digits of 582912 is $5+8+2+9+1+2=27$.