To solve this problem, we can use dynamic programming to find the maximum sum of the increasing subsequence.

1. Initialize an array dp of size n, where dp[i] represents the maximum sum of the increasing subsequence ending at index i.

2. Initialize dp[0] as arr[0], since the maximum sum of an increasing subsequence ending at the first element is the element itself.

3. Iterate from index 1 to n-1:
- For each index i, initialize dp[i] as arr[i].
- Iterate from index 0 to i-1:
- If arr[i] is greater than arr[j], update dp[i] as the maximum of dp[i] and dp[j] + arr[i]. This means that we can extend the increasing subsequence ending at index j by including the element at index i, if it is greater.

4. Find the maximum value in the dp array and return it as the result.

The time complexity of this approach is O(N^2) and the space complexity is O(N).

Here is the implementation of the maxSumIS() function in Python:

```python
def maxSumIS(N, arr):
dp = [0] * N
dp[0] = arr[0]

for i in range(1, N):
dp[i] = arr[i]
for j in range(i):
if arr[i] arr[j]:
dp[i] = max(dp[i], dp[j] + arr[i])

return max(dp)

# Example usage
N = 5
arr = [1, 101, 2, 3, 100]
print(maxSumIS(N, arr)) # Output: 106

N = 4
arr = [4, 1, 2, 3]
print(maxSumIS(N, arr)) # Output: 6
```

Note: The code provided above is in Python.

Question

To solve this problem, we can use dynamic programming to find the maximum sum of the increasing subsequence.

1. Initialize an array dp of size n, where dp[i] represents the maximum sum of the increasing subsequence ending at index i.

2. Initialize dp[0] as arr[0], since the maximum sum of an increasing subsequence ending at the first element is the element itself.

3. Iterate from index 1 to n-1:
   - For each index i, initialize dp[i] as arr[i].
   - Iterate from index 0 to i-1:
     - If arr[i] is greater than arr[j], update dp[i] as the maximum of dp[i] and dp[j] + arr[i]. This means that we can extend the increasing subsequence ending at index j by including the element at index i, if it is greater.
   
4. Find the maximum value in the dp array and return it as the result.

The time complexity of this approach is O(N^2) and the space complexity is O(N).

Here is the implementation of the maxSumIS() function in Python:

```python
def maxSumIS(N, arr):
    dp = [0] * N
    dp[0] = arr[0]
    
    for i in range(1, N):
        dp[i] = arr[i]
        for j in range(i):
            if arr[i] > arr[j]:
                dp[i] = max(dp[i], dp[j] + arr[i])
    
    return max(dp)

# Example usage
N = 5
arr = [1, 101, 2, 3, 100]
print(maxSumIS(N, arr))  # Output: 106

N = 4
arr = [4, 1, 2, 3]
print(maxSumIS(N, arr))  # Output: 6
```

Note: The code provided above is in Python.

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