After reading through metaMSE I'll answer the question myself, as the error in the original posting was quickly found, thanks again! And heavy editing of the original post destroys the context for the comments.
As $⌊250−50⋅x⌋\neq250−⌊50⋅x⌋$ contrary to what I wrote above, the derivation of the expectation proceeds as follows. We have $250-\lfloor50\cdot X\rfloor$ where $X\sim N(0, 1)$. Whenever $250-\lfloor50\cdot X\rfloor<0$ is true we forget the result and draw again. Due to
$$250<\left\lfloor50\cdot \frac{251}{50}\right\rfloor=251$$
we accept only draws within the half-open interval $X\in[-\infty, 251/50)$, consequently the normal distribution is truncated to this interval. The pdf of this truncated normal distribution is
$$
f(x)=\frac{g(x)}{\Phi(251/50)}\qquad g(x)=\begin{cases}0&x\ge \frac{251}{50}\\\phi(x)&x<\frac{251}{50}\end{cases}
$$
Integrating
$$
\int_{-\infty}^\infty f(x)dx=\frac{1}{\Phi(251/50)}\int_{-\infty}^\infty g(x)dx=\frac{1}{\Phi(251/50)}\int_{-\infty}^{\frac{251}{50}}\phi(x)dx=\frac{\Phi(251/50)}{\Phi(251/50)}=1
$$
confirms that we have a probability distribution. To calculate the expectation
$$\mathbb{E}\bigl[250-\lfloor50\cdot X\rfloor\bigr]=250-\int_{-\infty}^\infty\lfloor50\cdot x\rfloor f(x) dx
$$
we use the same approach as in the original posting and treat the floor function as what it is, a piecewise constant function. Thus creating a step function allows to collect all floating-point values that are mapped by the floor function to the same integer $a$. We have
$$
\begin{align}
250-\lfloor50\cdot x\rfloor&=a\in\mathbb{N}_0\\
250-a&=\lfloor50\cdot x\rfloor\Leftrightarrow x\in\left[\frac{250-a}{50},\frac{250-a+1}{50}\right)
\end{align}$$
Thus integrating over this interval
$$
h(a)=\int_{\frac{250-a}{50}}^{\frac{250-a+1}{50}}f(x)dx=\frac{1}{\Phi(251/50)}\int_{\frac{250-a}{50}}^{\frac{250-a+1}{50}}g(x)dx=\frac{\Phi\left(\frac{250-a+1}{50}\right)-\Phi\left(\frac{250-a}{50}\right)}{\Phi(251/50)}
$$
gives just the probability for ending up in the interval. Checking that we actually have a distribution is easy the as the sum nicely telescopes.
$$
\sum_{a=0}^{\infty}h(a)=\sum_{a=0}^{\infty}\frac{\Phi\left(\frac{250-a+1}{50}\right)-\Phi\left(\frac{250-a}{50}\right)}{\Phi(251/50)}=\frac{\sum_{a=0}^{\infty}\Phi\left(\frac{250-a+1}{50}\right)-\Phi\left(\frac{250-a}{50}\right)}{\Phi(251/50)}=\frac{\Phi\left(\frac{250-0+1}{50}\right)}{\Phi(251/50)}=1$$
Thus the expectation is simply
$$
\begin{align}
\sum_{a=0}^\infty a\cdot h(a)&=\sum_{a=1}^\infty a\cdot\frac{\Phi\left(\frac{250-a+1}{50}\right)-\Phi\left(\frac{250-a}{50}\right)}{\Phi(251/50)}\\
&=\frac{1}{\Phi(251/50)}\left(\sum_{a=1}^\infty a\cdot \Phi\left(\frac{251-a}{50}\right)-\sum_{a=1}^\infty a\cdot \Phi\left(\frac{250-a}{50}\right)\right)\\
&=\frac{1}{\Phi(251/50)}\left(\sum_{a=0}^\infty (a+1)\cdot \Phi\left(\frac{251-(a+1)}{50}\right)-\sum_{a=1}^\infty a\cdot \Phi\left(\frac{250-a}{50}\right)\right)\\
&=\frac{1}{\Phi(251/50)}\left(\Phi\left(\frac{250}{50}\right)+\sum_{a=1}^\infty (a+1)\cdot \Phi\left(\frac{250-a}{50}\right)-a\cdot \Phi\left(\frac{250-a}{50}\right)\right)\\
&=\frac{1}{\Phi(251/50)}\left(\Phi\left(\frac{250}{50}\right)+\sum_{a=1}^\infty \Phi\left(\frac{250-a}{50}\right)\right)\\
&=\frac{\sum_{a=0}^\infty \Phi\left(\frac{250-a}{50}\right)}{\Phi(251/50)}\approx250.500067...
\end{align}
$$
Any chance for a closed form of the sum? Probably not.
First. Assuming your year is between more than $28$ years away from a year divisible by $100$ by not divisible be $400$. (This will hold for the years $1829-1871, 1929-2071, 2129-2179$ etc.)
For these span of years every year with $28$ years before and $28$ years later, it will hold that every four years will be a leap year.
Non-leap years will have $365 = 52*7 + 1$ days so each consecutive year will normally start one day later than the next. However the year after a leap year will occur two days after the previous year.
If you compare year $n$ to year $n + k$ and and if there are $j$ leap years between $n$ and $k$ then the year will start $k + j$ days later.
Every $28$ years the entire calendar system starts over again because $28$ years will have $7$ leap years and $28 + 7 = 35 = 5*7$ so the calendar will start on the same day and will be a leap year if the first year was a leap year and won't be a leap year if the year wasn't a leap year.
So. Year $n$....
Case 1: Year $n$ is a leap year. The calendar will repeat in $28$ years and was the same $28$ years ago.
Case 2: Year $n$ is one year more than a leap year. $n+6$ will have one leap year between them ($n + 3$) and so $6 + 1 =7$ so calendar $n + 6$ will start on the same day and will not be a leap year so the calendars will be the same.
Year $n-5$ will be a leap year and not the same calendar. $n -6$ will have two leap years between them $(n-1, n-5)$ and will start $6+2 = 8$ earlier. $n-11$ will have three leap years between them ($n-1, n-5, n-9$) and so will start $11 + 3 =14 = 2*7$ days earlier and will be the same calendar.
Case 3: $n$ is two years past a leap year.
$n+5$ is not the same date because there is one leap year between them so the calendars or off by $5+1=6$ days. $n+6$ is not the same calendar. There is one leap year between the so $6+1 = 7$ and they start on the same day, but $n+6$ is a leap year. We must go further. $n+11$ will have $3$ leap years between them ($n+2, n+6,n+10$ and thus will start $11 + 3 = 14=2*7$ days later and will be the same calendar.
$n-5$ isn't the same. (One leap year and $5$ days isn't a multiple of $7$.) Nor is $n-6$ (it's a leap year). But $n-11$ will have three leap years $(n-2, n-6, n-10)$ and so will be $11 + 3 = 14$ days offset and the calendars will be the same.
Case 4: $n$ is 3 years past a leap year (like $2019$ is)
Then $n+5$ is a leap year $n+6$ has two leap years between and $n + 11$ will have $3$ leap years ($n+1, n+5, n+9$) and so be offset by $14$ and have the same calendar.
So $2030$ will be the next year with the same calendars.
And $n-6$ will have one leap year between them $n-3$ and so be offset by $6+1 = 7$ days and have the same calender. So $2013$ had the same.
Monkey Wrench. Years divisible by $100$ by not by $400$ do not have leap days and they throw the system off.
But again we can calculate those much the same.
Best Answer
I don't think that there's an easy way of doing this. Discrete RV's are messy ;0
However, your answer seems wrong. For example, suppose you add some integer (say, $100$) to X. Then the expected value should clearly also increase by $100$. But in your formula, say 1st one, $$(i_0 + i_1)/2 + i_0(0.5-y_0)/(x_1-x_0) + i_1(0.5+y_1)/(x_1-x_0)$$ $E(X)$ would increase by a higher amount due to multiplication by $i_0$ in 2nd term and $i_1$ in 3rd term.
Another way, if $y_0 = y_1 = 0$ then your answer should be $(i_0 + i_1)/2$.