Let the A.P. be $a_n=a_1+(n-1)d.$ We are given $a_1+(a_5+a_{10}+a_{15}+\cdots+a_{2020})+a_{2024}=2233.$

We need to find $S_{2024}=a_1+a_2+\cdots+a_{2024}.$

1. Write the given sum in terms of $a_1$ and $d$

The terms inside the bracket are $a_5, a_{10}, a_{15}, \ldots, a_{2020}.$ These are terms whose indices are multiples of $5$ from $5$ to $2020$.

Number of such terms: $\frac{2020}{5}=404.$

Now, $a_{5k}=a_1+(5k-1)d, \qquad k=1,2,\ldots,404.$

So, $\sum_{k=1}^{404} a_{5k} = \sum_{k=1}^{404}\bigl(a_1+(5k-1)d\bigr).$

Hence, $\sum_{k=1}^{404} a_{5k}=404a_1+d\sum_{k=1}^{404}(5k-1).$

Now, $\sum_{k=1}^{404}(5k-1)=5\sum_{k=1}^{404}k-404.$ Also, $\sum_{k=1}^{404}k=\frac{404\cdot 405}{2}=81810.$ Therefore, $5\sum_{k=1}^{404}k-404=5(81810)-404=409050-404=408646.$

Thus, $a_5+a_{10}+\cdots+a_{2020}=404a_1+408646d.$

Also, $a_{2024}=a_1+2023d.$

So the given condition becomes $a_1+(404a_1+408646d)+(a_1+2023d)=2233,$ that is, 406a_1+410669d=2233. \tag{1}

2. Express $S_{2024}$

Sum of first $2024$ terms of an A.P. is $S_{2024}=\frac{2024}{2}\bigl(2a_1+2023d\bigr)=1012(2a_1+2023d).$

So we need $2a_1+2023d$.

3. Relate equation (1) to $2a_1+2023d$

Observe that $410669=2023\times 203.$ Also, $406=2\times 203.$ Hence, $406a_1+410669d = 203(2a_1+2023d).$

From (1), $203(2a_1+2023d)=2233.$ So, $2a_1+2023d=\frac{2233}{203}=11.$

4. Compute the required sum

Therefore, $S_{2024}=1012\times 11=11132.$

5. Final answer

$a_1+a_2+\cdots+a_{2024}=11132.$