If N numbers add up to 2022, then the mean value of this set is 2022/N, and the first member of the sequence, "a", is 2022/N - (N-1)/2.

For the sequence elements to be integers, then either:

N is odd and 2022/N is an integer, or 

N is even and 2022/N is an integer+1/2

A programs yields these solutions, showing N, the first integer of the sequence and the last:

N first last

1 2022 2022

3 673 675

4 504 507

12 163 174

337 -162 174

1011 -503 507

1348 -672 675

4044 -2021 2022

A function below I called sumNstartwitha(a,N) computes the sum of a sequence of consecutive integers starting with "a" and containing N elements.  In each case this was equal to 2022.

-----------     code     -----------

listOf_a = []

listOf_N = []

for i in range(1,1000000):

    if i%2 == 0   and   (2022/i)%1 == .5:

        listOf_a.append(int((2022/i)  - (i-1)/2  ))

        listOf_N.append(i)

        print(i, int((2022/i)  - (i-1)/2  ), int((2022/i)  - (i-1)/2  ) + i - 1)

    if i%2 == 1   and   (2022/i)%1 == 0:

        listOf_a.append(int((2022/i)  - (i-1)/2  ))

        listOf_N.append(i)

        print(i, int((2022/i)  - (i-1)/2  ), int((2022/i)  - (i-1)/2  ) + i - 1)

Summing a, a+1, ... a+N-1

Let the 1st element of the sequence be "a".  Then the series to be summed is a, a+1, ... , a+N-1.

And this sum is a*N + sum(0,1,...,N-1) = a*N + (N)(N-1)/2 = (N)(2a+N-1)/2 = (N^2 + 2aN - N)/2

-----------   function   -----------

def sumNstartwitha(a,N):

    """  returns sum of N consecutive integers starting from a   """

    return  int(N*(2*a + N - 1)/2)