Every year, there is a contest...

First, you’re right in thinking that the total weight is $288.75$; you can also get this by noticing that each pumpkin is weighed four times, and $\frac{1155}4=288.75$. However, the data are inconsistent.

The low total of $108$ must be the sum of the two lightest weights, so the three heaviest pumpkins altogether weigh $288.75-108=180.75$. The two heaviest together weigh $122$, so the middle pumpkin weighs $180.75-122=58.75$. Each pair of pumpkins has an integer total weight, so each of the other four weights must have a fractional part of $0.25$ – but that ensures that the sum of any two of their weights is not an integer.

Observations

Let the five terms be represented by $a<b<c<d<e$.

Let's suppose there are solutions to the problem.

On the lower end, we know that $a+b=108$.

We also can deduce that $a+c=112$

On the upper end, we know that $d+e=122$

We can also deduce that $c+e=120$

Substituting, we discover that $c=b+4$ and $c=d-2$.

Now we know the difference between our central 3 values. The center should be ~$b+3$

$$b=b$$ $$c=b+4$$ $$d=b+6$$

The range is $6$. As you already pointed out, the mean is $57.75$, which we'll say is $58$ to make calculations easier.

Using this, let's guess that $b$ is $3$ below $58$. When this is the case, our values are the following:

$$a=53$$

$$b=55$$

$$c=59$$

$$d=61$$

$$e=61$$

To achieve $b,c,$ and $d$, simply plug in $57$ for $b$ in the above calculations and you'll obtain $a$ and $e$ by subtracting $b$ from $108$ and $d$ from $122$, respectively. Obviously these cannot be the correct weights since $d=e$, but let's add each combination to see how close we are. Upon doing so, we get the following values:

$$a+b=108$$

$$a+c=112$$

$$a+d=114$$

$$b+c=114$$

$$a+e=114$$

$$b+d=116$$

$$b+e=116$$

$$c+d=120$$

$$c+e=120$$

$$d+e=122$$

Now we'll try guessing that $b$ is $3.5$ less than $58$

In this case, we get the following:

$$a=53.5$$ $$b=54.5$$ $$c=58.5$$ $$d=60.5$$ $$e=61.5$$

These values yield the following:

$$a+b=108$$ $$a+c=112$$ $$b+c=113$$ $$a+d=114$$ $$a+e=115$$ $$b+d=115$$ $$b+e=116$$ $$c+d=116$$ $$c+e=120$$ $$d+e=122$$

This set was extremely close, but we had duplicate $115$ and $116$ values and lacked our $117$ and $118$ values.

I know others have already shown that there is no solution, but I thought this might help show that although it cannot be done, there are pumpkin weights that yield almost all of your values.