A man has a certain number of small boxes to pack into parcels. If he packs 3, 4, 5 or 6 in a parcel, he is left with one; if he packs 7 in a parcel, none is left over. What is the number of boxes he may have to pack?

Let the total number of boxes be $N$.
According to the question, when $N$ is divided by 3, 4, 5, or 6, the remainder is 1.
Therefore, $N$ can be written in the form:

$$N=\text{LCM}(3,4,5,6)\times k+1$$

where $k$ is a positive integer.

First, let's find the LCM of 3, 4, 5, and 6:

$$\text{LCM}(3,4,5,6)=60$$

So, the general form of the number is:

$$N=60k+1$$

We are also given that when the boxes are packed in groups of 7, no boxes are left over. This means $N$ must be perfectly divisible by 7:

$$(60k+1)(\mathrm{mod}7)=0$$

We can simplify $60(\mathrm{mod}7)$:

$$60=7\times 8+4⟹60\equiv 4(\mathrm{mod}7)$$

Substitute this back into the equation:

$$(4k+1)(\mathrm{mod}7)=0$$

Now, let's test integer values for $k$:

* For $k=1$: $4(1)+1=5$ (not divisible by 7)
* For $k=2$: $4(2)+1=9$ (not divisible by 7)
* For $k=3$: $4(3)+1=13$ (not divisible by 7)
* For $k=4$: $4(4)+1=17$ (not divisible by 7)
* For $k=5$: $4(5)+1=21$ (which is perfectly divisible by 7)

Since $k=5$ satisfies the condition, we can find $N$:

$$N=60\times 5+1=300+1=301$$

Thus, the number of boxes he may have to pack is 301.