The number of multiples of $k$ that are less than or equal to some integer $n$ is $\left\lfloor\frac{n}k\right\rfloor$, the largest integer $\le\frac{n}k$. For instance, the number of multiples of $3$ less than or equal to $10$ is $\left\lfloor\frac{10}3\right\rfloor=3$: they are $3,6$, and $9$. Thus, in your problem $N_3=\left\lfloor\frac{1000}3\right\rfloor=333$, $N_4=\left\lfloor\frac{1000}4\right\rfloor=250$, and so on. There's no problem with numbers near $1000$.
The calculation that you're trying to do is indeed an inclusion-exclusion calculation, but your notation needs to be fixed. If $N_3,N_4,N_5$, and $N_6$ are respectively the numbers of multiples of $3,4,5$, and $6$ less than or equal to $1000$, then you don't want $\sum_{i=1}^4N_i\;:$ that means $N_1+N_2+N_3+N_4$, and you don't even have an $N_1$ or $N_2$. You want $\sum_{i=3}^6N_i=N_3+N_4+N_5+N_6\;.$ This is standard summation notation.
The idea of the inclusion-exclusion calculation is that a first approximation to $N$, the number of multiples of $3,4,5$, or $6$ less than or equal to $1000$, can be obtained by adding together the number of multiples of $3$, of $4$, of $5$, and of $6$: $N\approx\sum_{i=3}^6N_i=N_3+N_4+N_5+N_6\;.\tag{1}$ Of course this is a very bad approximation, since these categories overlap a lot, as in your Venn diagram. For instance, $15$ is both a multiple of $3$ and a multiple of $5$, so $(1)$ counts it twice. Therefore we count the numbers that are multiples of both $3$ and $5$ and subtract that number to correct for the double-counting in $(1)$. Specifically, if $N_{j,k}$ is the number of positive integers less than or equal to $1000$ that are divisible by both $j$ and $k$, we want to subtract $N_{j,k}$ from the total in $(1)$. There are actually six such pairs; their counts are $N_{3,4},N_{3,5},N_{3,6},N_{4,5},N_{4,6}$, and $N_{5,6}$. Using summation notation we can represent the sum of these numbers as $\sum_{3\le j and we want to subtract this from $(1)$ to get a second, better approximation:
$N\approx\sum_{k=3}^6N_k-\sum_{3\le j
Unfortunately, it turns out that we've overcorrected: a number like $60$ that is a multiple of $3,4$, and $5$, say is counted $3$ times in $(1)$, and both counted and removed $(3)$ times in $(2)$, so it isn't counted at all in $(2)$.
To correct for this, we count the numbers that are multiples of three of the divisors $3,4,5$, and $6$ and add them back in. There are four such combinations, whose counts are $N_{3,4,5},N_{3,4,6},N_{3,5,6}$, and $N_{4,5,6}$. Their sum is
$\sum_{3\le i
and we get our third approximation by adding it back in:
$N\approx\sum_{k=3}^6N_k-\sum_{3\le j
Once again it turns out that we've overcorrected, so we subtract $N_{3,4,5,6}$, the number that are multiples of $3,4,5$, and $6$. At this stage we have an exact count:
$N=\sum_{k=3}^6N_k-\sum_{3\le j
Of course if we'd really been thinking, we could have ignored the multiples of $6$ from the start: every multiple of $6$ is already a multiple of $3$, so there's no point counting it separately. Thus, the inclusion-exclusion formula could have been reduced to $N=\Big(N_3+N_4+N_5\Big)-\Big(N_{3,4}+N_{3,5}+N_{4,5}\Big)-N_{3,4,5}\;.$