Calculate the lowest number where the sum of the sequence of numbers exceeds a given value

Jelly, 18 12 11 10 bytes

1Æs>¥#ḢṄÆs

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-1 byte thanks to Mr. Xcoder!

How it works

1Æs>¥#ḢṄÆs - Main link. Argument: n (integer)
1   ¥#     - Find the first n integers where...
 Æs        -   the divisor sum
   >       -   is greater than the input
       Ṅ   - Print...
      Ḣ    -   the first element
        Æs - then print the divisor sum

Brachylog, 9 bytes

∧;S?hf+S>

This program takes input from the "output variable" ., and outputs to the "input variable" ?. Try it online!

Explanation

∧;S?hf+S>
∧;S        There is a pair [N,S]
   ?       which equals the output
    h      such that its first element's
     f     factors'
      +    sum
       S   equals S,
        >  and is greater than the input.

The implicit variable N is enumerated in increasing order, so its lowest legal value is used for the output.

Japt, 15 bytes

[@<(V=Xâ x}a V]

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Explanation

Implicit input of integer U. [] is our array wrapper. For the first element, @ }a is a function that run continuously until it returns a truthy value, passing itself an incrementing integer (starting at 0) each time, and outputting the final value of that integer. â gets the divisors of the current integer (X), x sums them and that result is assigned to variable V. < checks if U is less than V. The second element in the array is then just V.