Finding pairs of Amicable Numbers, up to 10000
printf(&quot;|---------- PROGRAM FOR AMICABLE NUMBERS.----------|&quot;);

This kind of banner message makes it harder to use the output of your program in a pipeline.  I'd suggest removing it (the user has chosen to run it; trust them to know what they're doing!).

    int num1,num2,sum=0;

What's sum for?  It doesn't seem to be used.

    for(num1=1; num1&lt;=10000; num1++)

Where does the magic value 10000 come from?  It should be a named constant (or, better, specifiable as a command-line argument).

        for(num2=1; num2&lt;=10000; num2++)

Do we really need to consider every pair in both directions?  We could simply iterate while num2 &lt; num1 (which also saves ourselves a comparison later).  But see below why this loop isn't needed at all.

                if ((num1==sum_of_divisors(num2))

There's no prototype visible for sum_of_divisors() - enable the relevant compiler warnings and provide the prototype (e.g. by moving the definition ahead of main()).

        for(num2=1; num2&lt;=10000; num2++)
            {
                if ((num1==sum_of_divisors(num2))
                    &amp;&amp; (num2==sum_of_divisors(num1))
                    &amp;&amp; num1!=num2)

In the inner loop, sum_of_divisors(num1) is a constant, so we could save it in the outer loop, and make that the first comparison (so that short-circuit &amp;&amp; will then evaluate sum_of_divisors(num2) just once).  Then observe that we have a loop which tests whether its index is equal to a particular value - that means that we replace the inner loop with a single test:
    for (num1 = 1;  num1 &lt;= 10000;  num1++) {
         num2 = sum_of_divisors(num1);
         if (num1 &lt; num2 &amp;&amp; sum_of_divisors(num2) == num1) {
             /* we found a pair */
         }
    }

The num2 &lt; num1 test there saves us re-finding pairs we've already seen.  num2 &gt; num1 would also work, of course - I chose this version so that we print each pair lowest-first, as that seems to be the convention.

                    printf(&quot;\n%d\t\t%d&quot;, num1,num2);

It's better to consistently end your output with newline, rather than beginning with newline.  This works better in programs that might emit errors or other diagnostics, and plays nicely with line-buffered output such as interactive terminals (where the newline causes a flush).

    for(i=1; i&lt;n; i++) if(n%i==0) sum += i;

This is a really inefficient way to accumulate factors.  You can reduce it to the range (1, √n) with sum += i + n/i (adjusted a little to avoid double-counting √n when n is square); it may be better instead to generate all the prime factors and use those to generate the composite factors.

After making the above improvements, I found the runtime was below my measurement threshold.  Increasing the limit to 1 million gives a runtime of about 2 seconds - that might be acceptable; if not, the problem parallelizes well (other than outputting the results).

Modified version
#include &lt;stdio.h&gt;

#if USE_LONG_TYPE
typedef unsigned long Number;
#define FMT &quot;%lu&quot;
#else
typedef unsigned int Number;
#define FMT &quot;%u&quot;
#endif

static const Number MAX_VALUE = 1000000;


Number sum_of_factors(Number n)
{
    Number sum = 1;
    Number i = 2;
    for (;  i &lt; n/i;  ++i) {
        if (n/i*i == n) {
            sum += i + n/i;
        }
    }
    if (i*i == n)
        sum += i; /* add square root only once */
    return sum;
}

int main()
{
    //#pragma omp parallel for
    for (Number num1 = 1;  num1 &lt;= MAX_VALUE;  ++num1) {
        Number num2 = sum_of_factors(num1);
        if (num2 &gt; num1 &amp;&amp; num1 == sum_of_factors(num2)) {
            printf(FMT &quot;\t\t&quot; FMT &quot;\n&quot;, num1, num2);
        }
    }
}