Because $N > 0$, we need to have $x > y > 0$ for any solution. Since $x^2 - y^2 = (x + y)(x - y)$, each solution gives a factorisation of $N$ into two factors, say $N = a \cdot b$, where $a = x + y > x - y = b$.
If we start with a factorisation $N = a \cdot b$ with $a > b$, there is at most one solution $(x, y)$ corresponding to this factorisation of $N$, namely
$$x = \frac{a+b}{2} \qquad y = \frac{a-b}{2}.$$
The condition on $a, b$ for giving an integer solution $(x, y)$ is $a \equiv b \pmod 2$ so that $a + b$ and $a - b$ are both even. If $N$ is odd, any factorisation into two positive integers will give a solution. When $N$ is even, however, both factors must be even and $N = 4M$ for some integer $M$. Each factorisation of $M$ into two non-equal factors will then provide a solution.

To find all factorisations of $N$, we write $N = p_1^{a_1} p_2^{a_2} p_3^{a_3} \cdots p_k^{a_k}$ where all $a_i > 0$ and $p_1 < p_2 < \cdots < p_k$. It is well known that the number of positive divisors of $n$ is then equal to $\tau(N) = (a_1 + 1)(a_2 + 1)\cdots(a_k + 1)$. If $\tau(N)$ is even, which is the case precisely when $N$ is not a perfect square, the number of factorisations of $N$ into two factors is equal to $\tau(N)/2$. If $N$ is a square, we only get $(\tau(N) - 1)/2$ factorisations with $a > b$.
For the equation to have exactly 24 solutions, we need to use $N$ odd with $\tau(N) = 48$ or $\tau(N) = 49$, or $N = 4M$ with $\tau(M) = 48$ or $\tau(M) = 49$. There is no need to consider $N = 4M$ with odd $M$ since $M < 4M$ and we are looking for the smallest $N$.
Because $a_i > 0$ for all $i$, there are only two possibilities to obtain $\tau(N) = (a_1 + 1)(a_2 + 1)\cdots(a_k + 1) = 7^2 = 49$, namely $k = 1$ and $a_1 = 48$, or $k = 2$ and $a_1 = a_2 = 6$.

The smallest odd $N = p_1^{48}$ is $3^{48}$. The smallest $N = 4M$ is $2^{50}$ where we have used $M = 2^{48}$. Clearly, $3^{48} = 27^{16} > 16^{16} = 2^{64} > 2^{50}$.
The smallest odd $N = p_1^6 p_2^6$ is $N = 3^6 \cdot 5^6$. The smallest $N = 4M$ is $N = 2^8 \cdot 3^6$ with $M = 2^6 \cdot 3^6$. Note that $2^{50} = 2^8 \cdot 4^{21} > 2^8 \cdot 3^6$ and $3^6 \cdot 5^6 > 3^6 \cdot 4^4 = 2^8 \cdot 3^6$. Hence, the smallest $N$ we found so far is $N_0 = 2^8 \cdot 3^6$.

We now aim at finding all $N$ for which $\tau(N) = 48 = 3 \cdot 2^4$. Since $a_i + 1 \ge 2$ we can have at most $k=5$ different prime factors in $N$. We distinguish cases according to the number $k$. We will always assure that $a_1 \ge a_2 \ge \dots \ge a_k$, because this gives the smallest possible odd $N$ when $p_1 < p_2 < \dots < p_k$ are the $k$ smallest odd primes. When we allow $p_1 = 2$, we need to consider $N = 4M$, i.e. $N = p_1^{a_1+2} p_2^{a_2} p_3^{a_3} \cdots p_k^{a_k}$ with $p_1 = 2$.
**Case 1 ($k=5$):** The only option here is $a_1 = 2$ and $a_2 = a_3 = a_4 = a_5 = 1$, which leads to $N = p_1^2 p_2 p_3 p_4 p_5$ and the smallest odd number of this shape is $N_1 = 3^2 \cdot 5 \cdot 7 \cdot 11 \cdot 13$. The smallest even number $N = 4M$ we obtain here is $N_2 = 4 \cdot 2^2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 = 2^4 \cdot 3 \cdot 5 \cdot 7 \cdot 11$.

$$(5, 1, 1, 1) \quad (3, 2, 1, 1)$$
The smallest odd $N$ here are $N_3 = 3^5 \cdot 5 \cdot 7 \cdot 11$ and $N_5 = 3^3 \cdot 5^2 \cdot 7 \cdot 11$. The smallest even numbers are $N_4 = 2^7 \cdot 3 \cdot 5 \cdot 7$ and $N_6 = 2^5 \cdot 3^2 \cdot 5 \cdot 7$, respectively.

**Case 3 ($k=3$):** There are four possibilities for $(a_1, a_2, a_3)$:
$$(11, 1, 1) \quad (5, 3, 1) \quad (7, 2, 1) \quad (3, 3, 2)$$
This leads to the following eight possibilities for $N$:
$$\begin{array}{lll} (11, 1, 1) & N_7 = 3^{11} \cdot 5 \cdot 7 & N_8 = 2^{13} \cdot 3 \cdot 5 \\ (5, 3, 1) & N_9 = 3^5 \cdot 5^3 \cdot 7 & N_{10} = 2^7 \cdot 3^3 \cdot 5 \\ (7, 2, 1) & N_{11} = 3^7 \cdot 5^2 \cdot 7 & N_{12} = 2^9 \cdot 3^2 \cdot 5 \\ (3, 3, 2) & N_{13} = 3^3 \cdot 5^3 \cdot 7^2 & N_{14} = 2^5 \cdot 3^3 \cdot 5^2 \end{array}$$

**Case 4 ($k=2$):** There are four possibilities for $(a_1, a_2)$:
$$(23, 1) \quad (11, 3) \quad (7, 5) \quad (15, 2)$$
This leads to the following eight possibilities for $N$:
$$\begin{array}{lll} (23, 1) & N_{15} = 3^{23} \cdot 5 & N_{16} = 2^{25} \cdot 3 \\ (11, 3) & N_{17} = 3^{11} \cdot 5^3 & N_{18} = 2^{13} \cdot 3^3 \\ (7, 5) & N_{19} = 3^7 \cdot 5^5 & N_{20} = 2^9 \cdot 3^5 \\ (15, 2) & N_{21} = 3^{15} \cdot 5^2 & N_{22} = 2^{17} \cdot 3^2 \end{array}$$

Case 5 (k = 1): The only option is $a_1 = 23$ leading to odd $N_{23} = 3^{23}$ and even $N_{24} = 2^{25}$.

To find the smallest value of $N$, we first note that $2^{a+2} < 3^a$ for all $a \ge 4$. This is easily shown by induction. This implies that $N_{2m} < N_{2m-1}$ for all $m = 1, 2, \dots, 12$ except for $m = 1, 3, 7$. After cancelling common factors, it is easy to see that
$$\begin{align*} N_2 &= 2^4 \cdot 3 \cdot 5 \cdot 7 \cdot 11 < N_1 = 3^2 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \\ N_6 &= 2^5 \cdot 3^2 \cdot 5 \cdot 7 < N_5 = 3^3 \cdot 5^2 \cdot 7 \cdot 11 \\ N_{14} &= 2^5 \cdot 3^3 \cdot 5^2 < N_{13} = 3^3 \cdot 5^3 \cdot 7^2. \end{align*}$$
This shows that the smallest $N$ is one of the $N_{2m}$ where $m = 0, 1, 2, \dots, 12$. If we compare all $N_{2m}$ with $N_6$ we will find that $N_6 = 2^5 \cdot 3^2 \cdot 5 \cdot 7 = 10080$ is the smallest positive $N$ for which $x^2 - y^2 = N$ has exactly 24 solutions in positive integers.

To show $N_6 < N_{2m}$ in each case we cancel the common factors of these two numbers. To reduce our work, we may first observe that $3^2 \cdot 5 \cdot 7 = 316 < 2^9$, hence when $N_{2m}$ has a factor $2^a$ with $a \ge 14$ it is clear that $N_6 < N_{2m}$. Similarly, $N_6 < N_{2m}$ when $N_{2m}$ has a factor $2^a \cdot 3$ with $a \ge 12$, because $3 \cdot 5 \cdot 7 = 105 < 2^7$. Finally, since $5 \cdot 7 = 35 < 2^6$, whenever $N_{2m}$ has a factor $2^a \cdot 3^2$ with $a \ge 11$ we also have $N_6 < N_{2m}$. This last case can be extended to those $N_{2m}$ that have a factor $2^a \cdot 3^{2+i}$ with $a \ge 11-i$.