# [C (gcc)], <s>128</s> 125 bytes

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 p(_){printf("%d ",_);}f(n,x,y,i){x=n*n;y=1;for(i=0;++i<n;p(y),x-=y++)while(rand()&&(n-i)*(n-i+1)/2+(n-i)*(y+1)+y<x)y++;p(x);}

[Try it online!][TIO-jorc8zy5]

[C (gcc)]: https://gcc.gnu.org/
[TIO-jorc8zy5]: https://tio.run/##LY3RCsIwDEXf9xVFUJK1w00f675EhpS6acDFUae2jH177YZ5CPcecnNtcbM2xgEuOA2OeOxgs72Kjbqgnjtg5VVQhJOvOWcd6kp3TwdUl1pKOrEeIKDyRR2kxO@dHi04w1fA3Q64IMyXLSvcH@Tfh@RkOHlMiZT2qSamWtEbYliEcTerhL0bJ/KkP@cGxZSJNK/19Uh9CyWizlbYgRmfBOtl1Sx4oa4d345FqbM5xuMP "C (gcc) – Try It Online"

-3 bytes thanks to ceilingcat

NOTE: The probability is very very far from uniform. See explanation for what I mean and a better means to test that it works (by making the distribution closer to uniform [but still a far cry from it]).

### How?

The basic idea is to only choose increasing numbers so as to not worry about duplicates. Whenever we pick a number we have a non-zero chance of 'skipping' it if allowable. 

To decide if we can skip a number we need to know `x` the total left to be reached, `k` the number of elements we still have to use, and `y` the current candidate value. The smallest possible number that we could still pick would consist of $$y+(y+1)+(y+2)+...$$ added to the current value. In particular we require the above expression to be less than `x`. The formula will be $$\frac{k(k+1)}{2}+k(y+1)+y<x$$ Unfortunately we have to be a bit careful about rearranging this formula due to integer truncation in C, so I couldn't really find a way to golf it.

Nonetheless the logic is to have a chance to discard any `y` that satisfies the above equation.

### The code
 p(_){printf("%d ",_);} // Define print(int)
 f(n,x,y,i){ // Define f(n,...) as the function we want
 x=n*n; // Set x to n^2
 y=1; // Set y to 1
 for(i=0;++i<n;){ // n-1 times do...
 while(rand()&& // While rand() is non-zero [very very likely] AND
 (n-i)* // (n-i) is the 'k' in the formula
 (n-i+1)/2+ // This first half takes care of the increment
 (n-i)*(y+1) // This second half takes care of the y+1 starting point
 +y<x) // The +y takes care of the current value of y
 y++; // If rand() returned non-zero and we can skip y, do so
 p(y); // Print y
 x-=y++; // Subtract y from the total and increment it
 }p(x);} // Print what's left over.

The trick I mentioned to better test the code involves replacing `rand()&&` with `rand()%2&&` so that there is a 50-50 chance that any given y is skipped, rather than a 1 in `RAND_MAX` chance that any given y is used.