This problem requires calculating the determinant of the given matrix $M_j$, using the property of the adjugate matrix, and then evaluating the resulting summation.

Step 1: Use the property of the Adjugate Matrix
For an $n \times n$ matrix $M$, the determinant of its adjugate (or adjoint) matrix, adj(M), is given by the formula: \[ \operatorname{det}(\operatorname{adj}(M))=(\operatorname{det}(M))^{n-1} \] The matrix $M_j$ is a $3 \times 3$ matrix, so $n=3$. Therefore, the expression we need to evaluate simplifies to: \[ \sum_{j=0}^{24} \operatorname{det}\left(\operatorname{adj}\left(M_j\right)\right)=\sum_{j=0}^{24}\left(\operatorname{det}\left(M_j\right)\right)^{3-1}=\sum_{j=0}^{24}\left(\operatorname{det}\left(M_j\right)\right)^2 \]

Step 2: Calculate the Determinant of $M_j$
The matrix is $M_j=\left[\begin{array}{ccc}1 & 0 & \sqrt{2 j+1} \\ -1 & 1 & 0 \\ -\sqrt{2 j+1} & 0 & 1\end{array}\right]$. We will use cofactor expansion along the second column for simplicity, as it contains two zero entries: \[ \operatorname{det}\left(M_j\right)=(-1)^{2+2}(1) \cdot \operatorname{det}\left(\left[\begin{array}{cc} 1 & \sqrt{2 j+1} \\ -\sqrt{2 j+1} & 1 \end{array}\right]\right) \] The determinant of the $2 \times 2$ submatrix is: \[ \begin{gathered} \operatorname{det}\left(\left[\begin{array}{cc} 1 & \sqrt{2 j+1} \\ -\sqrt{2 j+1} & 1 \end{array}\right]\right)=(1)(1)-(\sqrt{2 j+1})(-\sqrt{2 j+1}) \\ =1-(-(2 j+1)) \\ =1+2 j+1 \\ =2 j+2 \end{gathered} \] So, the determinant of $M_j$ is: \[ \operatorname{det}\left(M_j\right)=2 j+2 \]

Step 3: Set up the Summation
Now we substitute $\operatorname{det}\left(M_j\right)$ back into the simplified summation expression: \[ \sum_{j=0}^{24}\left(\operatorname{det}\left(M_j\right)\right)^2=\sum_{j=0}^{24}(2 j+2)^2 \] We factor out $2^2=4$ : \[ \begin{gathered} \sum_{j=0}^{24}(2(j+1))^2=\sum_{j=0}^{24} 4(j+1)^2 \\ =4 \sum_{j=0}^{24}(j+1)^2 \end{gathered} \] Let $k=j+1$. When $j=0, k=1$. When $j=24, k=25$. The summation becomes: \[ 4 \sum_{k=1}^{25} k^2 \]

Step 4: Use the Formula for the Sum of Squares
We use the formula for the sum of the first $n$ squares, $\sum_{k=1}^n k^2=\frac{n(n+1)(2 n+1)}{6}$, with $n=25$ : \[ \begin{aligned} \sum_{k=1}^{25} k^2= & \frac{25(25+1)(2(25)+1)}{6} \\ & =\frac{25(26)(51)}{6} \end{aligned} \] We can simplify the fraction by canceling common factors: \[ \begin{gathered} =25 \cdot \frac{26}{2} \cdot \frac{51}{3} \\ =25 \cdot 13 \cdot 17 \\ =25 \cdot 221 \\ =5525 \end{gathered} \]

Step 5: Final Calculation
Substitute the sum back into the expression from Step 3: \[ \begin{gathered} \sum_{j=0}^{24} \operatorname{det}\left(\operatorname{adj}\left(M_j\right)\right)=4 \sum_{k=1}^{25} k^2 \\ =4 \times 5525 \\ =22100 \end{gathered} \] The value of $\sum_{j=0}^{24} \operatorname{det}\left(\operatorname{adj}\left(M_j\right)\right)$ is 22100.